I 


IN  MEMORIAM 

FLOR1AN  CAJORI 


PLANE   AND   SOLID   GEOMETRY 


PLANE    AND    SOLID 


GEOMETRY 


BY 


JAMES   HOWARD    GORE,   PH.D. 

n 

PROFESSOR  OF  MATHEMATICS,  COLUMBIAN  UNIVERSITY 

AUTHOR  OF  "ELEMENTS  OF  GEODESY,"  "HISTORY  OF  GEODESY," 

"  BIBLIOGRAPHY  OF _  GEODESY,"  ETC..,  ETC. 


•  • 


THIRD  EDITION  REVISED, 
WITH  AN  APPENDIX  OF  OVER  500  ADDITIONAL  EXERCISES. 


NEW  YORK 
LONGMANS,    GREEN,    AND    CO. 

91  AND  03  FIFTH  AVENUE,  NEW  YORK 
LONDON,    BOMBAY,    AND   CALCUTTA 

1908 


COPYRIGHT,  1898, 
BY  LONGMANS,   GREEN,   &   CO. 

COPYRIGHT,  1899, 
BY   LONGMANS,    GREEN,   &   CO. 


ALL   RIGHTS    RESERVED. 


CAJQRF 


«    t      . 


First  edition,  December,  1S98.     Reprinted  September,  1S99  ; 
March,  1902;   February,  1903;   August,  1904;   October,  1906 ; 
February,  1908. 


Nortoooti 

J.  S.  Gushing  Co.  —  Berwick  &  Smith  Co. 
Norwood,  Mass.,  U.S.A. 


INTRODUCTION. 


THE  study  of  Geometry  is  pursued  with  a  threefold  purpose, 

1.  To  aid  in  the  development  of  logical  reasoning. 

2.  To  stimulate  the  use  of  accurate  and  precise  forms  of 
expression. 

3.  To  acquire  facts  and  principles-  that  may  be  of  practical 
value  in  subsequent  life. 

The  first  two  purposes  are  advocated  because  of  their  dis- 
ciplinary importance ;  and  when  mathematics,  because  of  its 
exactness,  was  the  only  science  which  furthered  to  a  high 
degree  these  purposes,  it  was  necessary  for  the  student  to 
devote  a  large  part  of  his  time  to  their  study.  But  now  other 
sciences,  and  even  the  languages  and  philosophy,  claim  disci- 
plinary merit  equal  to  that  possessed  by  mathematics,  although 
differing  somewhat  in  the  character  of  the  training. 

Hence  it  appears  that  the  time  has  come  when  we  can  afford 
to  hearken  to  the  demands  of  the  utilitarians,  and  give  up 
those  refinements  in  mathematics  which  have  been  retained 
for  the  mental  discipline  they  bring  about,  but  which  are 
wholly  lacking  in  practical  application. 

I  have  therefore,  out  of  an  experience  as  a  computer  and 
worker  in  applied  mathematics,  as  well  as  a  teacher,  eliminated 
from  this  treatise  all  propositions  that  are  not  of  practical 
value  or  needed  in  the  demonstration  of  such  propositions. 


918306 


VI  INTRODUCTION. 


This  exclusion  leaves  out  about  one-half  of  the  matter  usually 
included  in  our  text-books  on  geometry.  However,  instructors 
will  not  entirely  miss  those  familiar  and  interesting  theorems 
which  helped  to  swell  the  books  they  studied,  —  such  theorems 
as  fall  below  the  practical  standard  are  here  given  as  exercises 
or  as  corollaries. 

Until  within  the  past  two  decades  the  verbiage  of  demon- 
strations was  so  elaborate  that  the  student  was  tempted  to 
memorize.  The  natural  reaction  resulted,  and  for  a  while  our 
authors  passed  to  the  other  extreme  in  symbolic  expressions. 
While  symbols  and  equational  statements  have  the  advantage 
of  brevity,  and  convey  information  to  the  mind  through  its 
most  receptive  channel,  -  -  the  eye,  —  still  they  discourage  the 
use  of  language,  and  hence  fail  to  develop  by  example  and 
precept  the  employment  of  accurate  and  precise  forms  of 
expression. 

I  have  therefore  sought  to  use  symbols  and  equations  only 
in  those  cases  where  I  could  see  no  gain  in  spelling  out  their 
meaning. 

Attention  is  called  to  the  solution  of  problems.  Ordinarily 
the  problem  is  presumed  to  be  solved,  and  then  a  demonstra- 
tion is  given  to  show  that  the  solution  was  correct.  This  does 
not  appear  to  me  to  be  in  the  line  of  discovery.  I  have  in  all 
cases  started  with  a  statement  of  those  known  facts  which 
plainly  suggest  the  first  step  in  the  solution,  then  introduced 
the  next  step,  giving  the  construction  in  connection  with 
each  stated  fact,  so  that  with  the  completed  construction  goes 
its  own  demonstration  and  the  student  sees  the  road  along 
which  he  travelled,  and  understands  from  the  beginning  why 
he  started  upon  it. 


INTI!  OD  UCTION.  vii 


Great  care  has  been  exercised  in  the  selection  of  exercises  to 
follow  each  demonstration.  They  are  intended  to  be  varia- 
tions upon  the  theorem  demonstrated,  or  extensions  of  it,  so 
that  at  least  a  portion  of  the  required  proof  is  suggested.  At 
the  end  of  each  book  will  be  found  a  larger  collection  of  exer- 
cises, formulae,  and  numerical  examples. 

In  conclusion,  I  may  state  that  no  claim  is  made  to  origi- 
nality in  demonstration ;  I  have  employed  those  I  deemed 
best ;  however,  no  statement  is  taken  from  another  author 
unless  it  is  the  common  property  of  several. 


TABLE   OF   CONTENTS. 


Preliminary  Definitions  . 
Postulates  ... 
Axioms  .... 
Abbreviations  ... 


PAGE 

1 
3 
3 
4 


PLANE    GEOMETRY. 


BOOK  I.     RECTILINEAR  FIGURES. 


Definitions  and  General  Principles 
Parallel  Lines  ..... 
Triangles.         .... 
Quadrilaterals  .... 
Polygons  ..... 


. 


17 
22 
33 
40 


BOOK  II.     THE  CIRCLE. 


Definitions        ... 
On  Measurement     .. 
Method  of  Limits     .. 
Measurement  of  Angles 
Problems  in  Construction 


46 
56 
57 
59 
65 


BOOK  III.     RATIO  AND  PROPORTION.     SIMILAR  FIGURES. 

Definitions        ...........     78 

Proportional  Lines  ..........     84 

Similar  Polygons      .  .      .        .         .         .         .        .        .         .        .87 

ix 


CONTENTS. 


BOOK  IV.     AREAS  OF  POLYGONS. 

PAGE 

Definitions 100 

Problems  in  Construction 118 

BOOK  V.     REGULAR  POLYGONS  AND  CIRCLES. 

Definitions- 124 

Maxima  and  Minima 139 


SOLID    GEOMETRY. 
BOOK  VI.     PLANES  AND  SOLID  ANGLES. 

Definitions 145 

Diedral  Angles 155 

Polyedral  Angles 159 

BOOK    VII.      POLYEDRONS,    CYLINDERS,    AND    CONES. 

General  Definitions 164 

Prisms  and  Parallelepipeds 164 

Pyramids        .  178 

BOOK  VIII.     THE  SPHERE. 

Definitions 192 

Spherical  Angles  and  Polygons 194 

The  Sphere 204 

Formulae         .         .  209 


Appendix  of  Additional  Exercises 211 


GEOMETRY 


PRELIMINARY   DEFINITIONS. 

1.  Space  has  extension  in  all  directions,  and  so  far  as  our 
experience  can  teach  us  it  is  limitless. 

2.  A  material  or  physical  body  occupies  a  definite  portion 
of  space,  and  this  space  freed  from  the  body  is  called  a  geomet- 
rical solid,  which  for  brevity  will  be  known  as  a  Solid. 

3.  The  limits,  or  boundaries,  of  a  solid  are  Surfaces. 
The  limits,  or  boundaries,  of  a  surface  are  Lines. 
The  intersection  of  two  lines  is  a  Point. 

4.  It  is  said  a  solid  has  three  dimensions :  Length,  Breadth, 
and  Thickness. 

A  surface  has  only  two  dimensions :  length  and  breadth. 

A  line  has  only  one  dimension :  length. 

A  point  is  without  dimension,  having  simply  position. 

5.  In  drawings  and  diagrams  material  figures  are  employed 
for  purposes  of  demonstration,  but  they  are  merely  the  repre- 
sentatives of  mathematical  figures. 

6.  A  Straight  Line,  or  Right  Line,  is  the  shortest  line  between 
two  points  ;  as  AB  A—  —B 

7.  A  Broken  Line  is  a  line  com- 
posed of  different  successive  straight 
lines ;  as  ABCDEF. 

8.  A    Curved   Line,  or    simply    a 
Curve,  is  a  line  no  portion  of  which 
is  straight;  as  ABC. 

1 


GEOMETR  Y. 


9.  A  Plane  Surface,  or  simply  a  Plane,  is  one  such  that  the 
straight  line  which  joins  any  two  of  its  points  lies  entirely  in 
the  surface. 

10.  A  Curved  Surface  is  one  no  portion  of  which  is  plane. 

11.  A  Geometrical  Figure  is  any  combination  of  points,  lines, 
surfaces,  or  solids  formed  under  specific  conditions. 

Plane  Figures  are  formed  by  points  and  lines  in  a  plane ;  Rec- 
tilinear, or  Jiitjht-Uiied  Figures,  are  formed  of  straight  lines. 

12.  Geometry  is  that  branch  of  mathematics  which  treats  of 
the  construction  of  figures,  of  their  measurement,  and  of  their 
properties. 

Plane  Geometry  treats  of  plane  figures. 

Solid  Geometry,  sometimes  called  Geometry  of  Space  and 
Geometry  of  Three  Dimensions,  treats  of  solids,  of  curved  sur- 
faces, and  of  all  figures  that  are  not  represented  on  a  plane. 

13.  A  Theorem  is  a  truth  requiring  demonstration. 

14.  A  Problem  is  a  question  proposed  for  solution. 

15.  A  Postulate  assumes  the  possibility  of  the  solution  of 
some  problem. 

16.  An  Axiom  is  a  truth  assumed  to  be  true,  or  a  truth  veri- 
fied by  intuition  or  our  experience  with  material  things. 

17.  A  Proposition  is  a  general  term    for  theorem,  axiom, 
problem,  and  postulate. 

18.  A  Demonstration  is  the  course  of  reasoning  by  which 
the  truth  of  a  theorem  is  established. 

19.  A  Corollary  is  a  conclusion  which  follows  immediately 
from  a  theorem,   but  this  conclusion  may  at  times  demand 
demonstration. 


AXIOM*. 


20.  A  Lemma  is  an  auxiliary  theorem  required  in  the  demon- 
stration of  a  principal  theorem. 

21.  A  Scholium  is  a  remark  upon  one  or  more  propositions. 

22.  An  Hypothesis  is  a  supposition  made  either  in  the  enun- 
ciation of  a  proposition  or  in  the  course  of  a  demonstration. 

23.  A  Solution  of  a  problem  is  the  method  of  construction 
which  accomplishes  the  required  end. 

24.  A  Construction  is  the  drawing  of  such  lines  and  curves 
as  may  be  required  to  prove  the  truth  of  a  theorem,  or  to  solve 
a  problem. 

25.  The  Enunciation  of  a  theorem  consists  of  two  parts  :  the 
Hypothesis,  or  that  which  is  assumed;  and  the  Conclusion,  or 
that  which  is  asserted  to  follow  therefrom. 

POSTULATES. 

26.  1.  A  straight  line  can  be  drawn  between  any  two  points. 
2.    A  straight  line  can  be  produced  indefinitely  in  either 

direction. 

27.  Given  Prove 


Axiom 
(Assumed  truth) 

Postulate 

(Assumed  possibility) 


Proposition 


• 


f  Theorem 
That  something  is  true 

Problem 

That  something  can  be 


done 
AXIOMS. 


28.  1.  Things  which  are  equal  to  the  same  thing  are  equal 
to  each  other. 

2.  If  equals  be  multiplied  or  divided  by  equals,  the  results 
will  be  equal. 


GEOMETRY. 


3.  If  equals  be  added  to  or   subtracted   from  equals,  the 
results  will  be  equal. 

4.  If  equals  are  added  to  or  subtracted  from  unequals,  the 
results  will  be  unequal. 

5.  The  whole  is  equal  to  the  sum  of  its  parts. 

6.  The  whole  is  greater  than  any  of  its  parts. 

ABBREVIATIONS. 

29.  The  following  is  a  list  of  the  symbols  which  will  be 
used  as  abbreviations : 

+  ,  plus.  >,  is  greater  than. 

— ,  minus.  <,  is  less  than. 

X ,  multiplied  by.  .*.,  therefore. 

=  ,  equals.  Z,  angle. 

In  addition  to  these,  the  following  may  be  used  for  writing 
demonstrations  on  the  board  or  in  exercise  books,  but  no  use  is 
made  of  them  in  the  present  work. 

[R,  right  angle.  A,  triangles. 

A,  angles.  O,  parallelogram. 

J_,  perpendicular.  /U,  parallelograms. 

Js,  perpendiculars.  O,  circle. 

||,  parallel.  CD,  circles. 

Us,  parallels.  o,  arc. 
A,  triangle. 


PLANE  GEOMETRY 

BOOK   I. 

RECTILINEAR  FIGURES. 

30.  An  Angle  is  the  difference  in  direction  of  two  lines  that 
meet  or  might  meet;  if  the  two  lines  meet,  the  point  of  meet- 
ing is  called  the  Vertex,  and  the  lines 

are  called  its  Sides.  Thus,  in  the  angle 
formed  by  AB  and  EC,  B  is  the  vertex, 
and  AB  and  BC  are  the  sides. 

31.  An  isolated  angle  may  be  des- 
ignated by  the  letter  at  its  vertex,  as 

"  the  angle  0 " ;  but  when  several  angles  are  formed  at  the 

same  point  by  different  lines,  as  OA,  OB,  OC,  we  designate 

the  angle  intended  by  three  letters ;  namely, 

by  one  letter  on  each  of  its  sides,  together 

with  the  one  at   its  vertex,  which  must  be 

written  between  the  other  two.     Thus,  with 

these  lines  there  are  formed  three  different 

angles,  which  are  distinguished  as  AOB,  BOO,  and  AOC. 

32.  Two  angles,  such  as  AOB,  BOC,  which  have  the  same 
vertex  0  and  a  common  side  OB  between  them,  are  called 
Adjacent. 

33.  The  magnitude  of  an  angle  depends  wholly  upon  the 
amount  of  divergence  of  its  sides,  and  is  independent  of  their 
length. 

5 


PLANE  GEOMETRY. 


[BK.  I. 


34.  Two  angles  are  Equal  when  one  can  be  placed  upon  the 
other  so  that  they  shall  coincide.  Thus,  the 
angles  AOB  and  A'O'B'  are  equal,  if  A'O'B' 
can  be  superposed  upon  AOB  so  that  while 
O'A'  coincides  with  OA,  O'B'  shall  also  coin- 
cide with  OB,  or  when  the  difference  in  the 

• 

directions  of  the  sides  of  one  angle  is  the 


BA 


B' 


same  as  the  difference  in  the  directions  of  the  sides  of  the  other. 

35.  When  one   straight   line   meeting   another   makes   the 
adjacent  angles  equal  to  each  other,  ^ 

each  of  the  angles  is  called  a  Right 
Angle;  and  the  two  lines  thus  meeting 
are  said  to  be  perpendicular  to  each 

other  or  at  right  angles  to  each  other.    ^ 

Thus,  if  the  adjacent  angles  AOC  and  0 

BOC  are  equal  to  each  other,  each  is  a  right  angle,  and  the 

line  CO  is  perpendicular  to  AB,  and  AB  is  perpendicular  to 

CO.     The  point  0  is  called  the  Foot  of 

the  perpendicular. 

36.  An  Oblique  Angle  is  formed  by 

one  straight  line  meeting  another  so  as         B Z C 

to  make  the  adjacent  angles  Unequal. 

Oblique  angles  are  subdivided  into  two  classes,  Acute  Angles 
and  Obtuse  Angles. 

37.  An  Acute  Angle  is  less  than  a 
right  angle,  as  the  angle  0.  o 

38.  An  Obtuse  Angle  is  greater  than  a  right  angle,  as  the 
angle  AOB  (in  36). 

39.  A  Straight  Angle  has  its  sides 
extending  in  opposite  directions  so  as 

to  be  in  the  same  straight  line.     Thus,    A 

if   OA,  OB  are  in  the   same   straight 

line,  the  angle  formed  by  them  is  called  a  straight  angle. 


O 


§  43.] 


RECTILINEAR  FIGURES. 


40.  The  Complement  of  an  angle  is  the 
difference  between  a  right  angle  and  the 
given  angle.  Thus,  ABD  is  the  comple- 
ment of  the  angle  DEC;  also  DBC  is  the 
complement  of  the  angle  ABD. 


B 


41.  The  Supplement  of  an  angle  is  the 
difference  between   a  straight   angle   and 
the  given  angle.     Thus,  ACD  is  the  sup- 
plement of  the  angle  DCB;  also  DCB  is 
the  supplement  of  the  angle  ACD. 

42.  Vertical   Angles    are   angles    which 
have  the  same  vertex,  and  their  sides  ex- 
tending in  opposite  directions.     Thus  the 
angles  AOD  and  COB  are  vertical  angles, 
as  also  the  angles  AOC  and  DOB. 

43.  The  magnitude  of  an  angle  is  meas- 
ured by  finding  the  number  of  times  which  it  contains  another 
angle  adopted  arbitrarily  as  the  unit  of  measurement. 

The  usual  unit  of  measurement  is  the  Degree,  or  the  ninetieth 
part  of  a  right  angle.  To  express  fractional  parts  of  the  unit, 
the  degree  is  divided  into  sixty  equal  parts,  called  Minutes,  and 
the  minute  into  sixty  equal  parts,  called  Seconds. 

Degrees,  minutes,  and  seconds  are  denoted  by  the  symbols 
°,  ',  ",  respectively;  thus,  28°  42'  36"  stands  for  28  degrees  42 
minutes  and  36  seconds. 


EXERCISES. 

1.  How  many  degrees  are  there  in  the  complement  of  27°  ?  of  65°  ? 
of  18°  IT  ?  of  38°  18'  35"  ?  of  f  of  a  right  angle  ? 

2.  How  many  degrees  are  there  in  the  supplement  of  68°  ?  of  124°  16'  ? 
of  142°  18'  46"  ?  of  f  of  a  right  angle  ? 

3.  How  many  degrees  are  there  in  an  angle  which  is  the  complement 
of  three  times  itself  ? 


8 


PLANE  GEOMETEY. 


[BK.  I. 


4.  How  many  degrees  are  there  in  an  angle  whose  supplement  is  two 
times  its  complement  ? 

5.  How  many  degrees  are  there  in  an  angle  if  its  complement  and 
supplement  are  together  equal  to  120°  ? 


PROPOSITION  I.     THEOREM. 

44.  If  a  straight  line  meets  another  straight  line,  the  sum  of 
the  adjacent  angles  is  equal  to  two  right  angles. 

Let  DC  meet  AB  at  C;  then  the  sum  of  the  angles  DC  A 
and  DCB  is  equal  to  two  right  angles. 

At  C,  let  CE  be  drawn  perpendicular 
to  AB\  then,  by  definition,  the  angles 
EGA  and  ECB  are  both  right  angles, 
and  consequently  their  sum  is  equal  to 
two  right  angles. 

The  angle  DC  A  is  equal  to  the  sum 
of  the  angles  EGA  and  ECD ;  hence, 

Z  DCB  +  Z  DC  A  =  (Z  DCB  +  Z  DOE)  +  Z  EGA 

=  Z  ECB  +  Z  EC  A. 

But  by  construction  Z  ECB  and  Z  1£(L!  are  right  angles, 

therefore 

Z  .DOT  +  Z  DC  A  =  2  right  angles. 

45.  COR.  1.   If  one  of  the  angles  DC  A,  DCB,  is  a  right  angle, 
the  other  must  also  be  a  right  angle. 

46.  COR.  2.    The  sum  of  the  angles 
BAG,  CAD,  DAE,  EAF,  formed  about 
a  given  point  on  the  same  side  of  a 
straight  line  BF,  is  equal  to  two  right  B 
angles.     For  their  sum  is  equal  to  the 

sum  of  the  angles  EAB  and  EAF;  which,  from  the  proposition 
just  demonstrated,  is  equal  to  two  right  angles. 


§48.]  RECTILINEAR   FIGURES.  9 

47.    COR.  3.    At  a  given  point  in  a  straight  line,  and  on  a 
given  side  of  the  line,  only  one  per- 
pendicular  to  that  line  can  be  erected.  D 

For  if  two  could  be  erected,  let  them 
be  EC  and  DC',  then  (by  35)  Z  ECB 
is  a  right  angle,  likewise  Z  DCB  is 
a  right  angle,  or  (by  28)  4 

/.ECB=/.DCB,  C 

which  is  impossible,  as  Z  DCB  is  a  part  of  Z  ECB. 

EXERCISE. 

If  in  Cor.  2  the  angles  EAF,  CAD,  and  BAG  are  equal,  and  each 
twice  as  large  as  the  angle  DAE,  what  will  be  the  size  of  each  angle  in 
degrees  ? 


PROPOSITION  II.      THEOREM. 

48.  CONVERSELY,*  if  the  sum  of  two  adjacent  angles  is  equal 
to  two  right  angles  or  to  a  straight  angle,  their  exterior  sides  lie  in 
the  same  straight  line. 


Let  the  sum  of  the  adjacent  angles  ACD  and  BCD  be  equal 
to  two  right  angles. 

To  prove  that  ACB  is  a  straight  line. 

If  ACB  is  not  a  straight  line,  let  CE  be  in  the  same  straight 
line  with  AC. 

*  Hereafter  converse  propositions  will  not  be  demonstrated,  but  given 
as  corollaries  or  exercises.  The  method  of  demonstration,  which  in  gen- 
eral will  be  identical  to  the  direct  proposition,  or  a  proof  that  any  other 
condition  would  not  be  true,  will,  however,  be  indicated. 


10  PLANE  GEOMETRY.  [BK.  I. 

Then  (by  41)  Z  ECD  is  the  supplement  of  Z  A  CD. 
But  by  hypothesis  Z  BCD  is  the  supplement  of  Z  ACD. 


E 


A  C  B 

Therefore,  since  supplements  of  the  same  angle  must  be 
equal  to  one  another,  Z  ECD  must  be  equal  to  Z  BCD,  which 
(by  28)  is  impossible  except  when  CE  coincides  with  CB,  or 
CB  is  the  only  line  that  is  a  prolongation  of  AC. 

PROPOSITION  III.     THEOREM. 

49.  If  two  straight  lines  intersect  each  other,  the  vertical  angles 
are  equal. 

.C 


D 

Let  the  straight  lines  AB  and  CD  intersect  at  E. 

To  prove  that  Z  AEG  =  Z  BED. 

By  (44),  Z  AEG  +  Z  CEB  =  two  right  angles, 
and  Z  BED  +  Z  CEB  =  two  right  angles. 

Therefore,  by  (28), 

Z  AEG  +  Z  OE£  -  Z  BED  +  Z  (7^5. 

Subtracting  Z  0^5  =  Z  GEB, 

we  have  Z  A^C  =  Z  J3#D.  Q.E.B, 

In  the  same  way  it  may  be  proved  that 

Z  AMD  =  Z  CEB. 


§  51.]  RECTILINEAR   FIGURES.  11 

50.  COR.  1.  If  two  straight  lines  intersect  each  other,  the  four 
angles  which  they  make  at  the  point  of  intersection  are  together 
equal  to  four  right  angles. 

If  one  of  the  four  angles  is  a  right  angle,  the  other  three  are 
right  angles,  and  the  lines  are  mutually  perpendicular  to  each 
other. 

51.  COR.  2.  If  any  number  of  straight  lines  meet  at  a  point, 
the  sum  of  all  the  angles  having  this  vertex  in  common  is  equal  to 
four  right  angles. 

The  sum  of  all  the  angles  AOB,  BOO, 
COD,  DO  A,  formed  about  a  point,  is  equal 
to  four  right  angles. 

For  if  the  line  OA  is  produced  to  E,  the  E~ 
sum  of  the  angles  AOB,  BOO,  and  COE  is 
equal  to  two  right  angles,  and  the  same  is 
true  of  the  sum  of  the  angles  AOD  and 
DOE. 

Hence  the  sum  of  the  angles  AOB,  BOC,  COD,  and  DOA  is 
equal  to  four  right  angles. 

EXERCISES. 

1.  If  in  the  above  figure  the  angles  AOB,  BOC,  and  AOD  are  respec- 
tively 42°,  85°,  and  |  of  a  right  angle,  how  many  degrees  are  there  in 
COD? 

2.  If  in  Prop.  III.  the  angle  CEA  is  34°  21',  how  many  degrees  are 
there  in  each  of  the  other  angles  ? 

3.  If  J.'s  complement  is  equal  to  one-sixth  of  A's  supplement,  find  A. 

4.  In  Prop.  III.  the  angle  CEA  is  equal  to  one-fourth  of  angle  CEB. 
How  many  degrees  are  there  in  each  of  the  other  angles  ? 

5.  If  in  Cor.  2,  angle  COE  is  equal  to  angle  EOD,  show  that  angle 
CO  A  is  equal  to  angle  AOD. 

6.  If  the  angles  BOA,  EOC,  and   COB  are  in  the  ratio  of  2  : 3  :  5, 
how  many  degrees  are  there  in  each  ? 

7.  If  the  angles  BOA,  BOC,  AOD,  and  COD  are   in   the  ratio  of 
1:2:3:4,  how  many  degrees  are  there  in  each  ? 


12 


PLANE  GEOMETRY. 


[BK.  I. 


PROPOSITION  IV.     THEOREM. 

52.  From  a  point  outside  a  straight  line  only  one  perpendicular 
can  be  drawn  to  such  straight  line,  and  this  perpendicular  is  the 
shortest  distance  from  the  point  to  the  line. 

Let  P  be  the  point,  AB  the  line, 
and  PO  a  perpendicular. 

1.  To  prove  that  PO  is  the  only 
perpendicular  that  can  be  drawn,  and 
that  it  is  the  shortest  distance  from 
P  to  AB. 

Produce  PO  to  P,  making  OP 
=  OP;  then  the  angles  FOB  and 
P'OB  are  right  angles. 

If  any  other  perpendicular  can  be 
drawn,  suppose  it  be  PQ,  and  join 

P'Q. 

Eevolve  the  figure  OPQ,  about  AB  as  an  axis ;  then,  since 
POQ  is  a  right  angle,  OP  will  fall  on  OP,  and  the  point  P  will 
coincide  with  P',  and,  Q  remaining  stationary,  PQ  will  fall 
upon  P'Q. 

Therefore,  if  Z  PQO  be  a  right  angle,  Z  P'QO  must  also  be 
a  right  angle,  and  (from  48)  the  lines  PQP'  must  be  straight ; 
this  would  give  two  straight  lines  joining  P  and  P',  which  is 
impossible,  or  PQ  cannot  be  perpendicular  to  AB. 

Hence  only  one  perpendicular  can  be  drawn. 

2.  To  prove  that  PO  is  the  shortest  distance  from  P  to  the 
line  AB. 

It  was  just  shown  that  OP  could  be  made  to  coincide  with 
OP',  and  PQ  with  QP',  or, 

PO=P'OandPQ=  QP'. 

Since  PP'  is  a  straight  line,  it  is  the  shortest  line  that  can 
fie  drawn  from  P  to  P', 


§54.]  RECT1LISEAE   FIGUKZ  -  13 

or  PP  <  PQP  • 

that  is,  PO^OP  <PQ+  QP', 
or  2  PO  <  2  PQ. 

Dividing  by  2,  PO  <  PQ. 

Hence  the  perpendicular  is  the  shortest  distance  from  a 
point  to  a  straight  line.  Q.E.D. 

PROPOSITIOX  V.     THEOREM. 

53.  T":o  oblique  lines  draicn  from  a  point  to  a  straight  line, 
cutting  off  equal  distances  from  the  foot  of  the  perpendicular,  are 
equal 

C 


A         D        B 

Let  the  oblique  lines  CA  and  CB  meet  the  line  EF  at  equal 
distances  from  the  foot  of  the  perpendicular  CD. 

To  prove  that  CA  =  CB. 

Let  the  part  CD  A  be  revolved  about  CD  until  DE  falls 
upon  its  prolongation,  DF\  then,  since  AD  —  DB  by  hypothe- 
sis, and  CD  remains  stationary,  the  point  A  will  fall  on  B,  and 
the  line  CA  will  fall  on  CB,  and  be  equal  to  it. 

Hence  CA  =  CB.  Q.E.D. 

54.  COR.  1.  Since  D  is  the  middle  point  of  AB,  DC  a 
perpendicular,  and  C  any  point  on  this  perpendicular,  it  is 
true  that  every  point  on  the  perpendicular  bisector  of  a  straight 
line  is  equally  distant  from  the  extremities  of  that  line, 


PLANE   GEOMETRY. 


[BK.  I. 


55.    COR.  2.    When  CAD  was  revolved,  it  was  found   that 
AD  fell  upon  BD  and  AC  upon  BC. 


and.  similarly, 


that  is,  the  two  equal  oblique  lines  drawn  from  the  same  point 
in  the  perpendicular  make  equal  angles  with  the  perpendicular 
and  also  with  the  base. 

56.  COR.  3.  If  two  2ioints  on  a  line  are  equally  distant  from 
the  extremities  of  another  line,  the  first  line  is  a  perpendicular 
bisector  of  the  second. 


PROPOSITION  VI.     THEOREM. 

57.  If  two  lines  are  drawn  from  a  point  to  the  extremities  of  a 
straight  line,  their  sum  is  greater  than  the  sum  of  two  other  lines 
similarly  drawn,  but  enveloped  by  them. 


E 


Let  AB  and  AC  be  drawn  from  the  point  A  to  the  extremities 
of  the  line  BC,  and  let  DB  and  DC  be  two  lines  similarly 
drawn,  but  enveloped  by  AB  and  AC. 


§58  a.] 


RECTILINEAR   FIGURES. 


15 


To  prove  that       AB  +  AC  >  DB  -f-  DC. 

Produce  BD  to  meet  AC  at  E. 
Since  BE  is  a  straight  line, 

BE  <  BA  +  AE. 
Add  #<?,  £#  +  EC  <  BA  +  AE  + 

<  £.4  +  AC. 
Since  Z>(7  is  a  straight  line, 


Add 


But 


Q.E.D. 


BD  +  DC<  (BD  +  DE)  + 

<BE  +  EC. 
\-EC<BA^ 


.-.  AB  +  AC  >  DB  +  DC. 


PROPOSITION  VII.     THEOREM. 

58.  Of  two  oblique  lines  draivn  from  the  same  point  to  the  same 
straight  line,  that  which  meets  the  line  at  the  greater  distance  from 
the  foot  of  the  perpendicular  is  the  greater. 

Let   PC  be   perpendicular  to  AB,  and   PD  and  PE  two 
oblique   lines  cutting  off  unequal  dis- 
tances  from  C. 

To  prove  PE  >  PD. 

Produce  PC  to  P',  making 


join  P'D  and  P'E  ; 

then  (by  54)  P'D  =  PD,  and  P'E=PE. 

By  (57)         PE  +  P'E>PD  +  P' 
or  2PE>2  PD. 

Dividing  by  2,  PE  >  PD. 


E  \  \D  C  D' 

\\ 


\\ 


\ 


Q.E.D. 


58  a.   COR.    Only  two  equal  straight  lines  can  be  drawn  from  a 
point  to  a  straight  line. 


16 


PLANE  GEOMETRY. 


[BK.  I. 


D. 


EXERCISES. 

1.  If  the  oblique  lines  are 
on  opposite  sides  of  the  per- 
pendicular,   show     that     the 
theorem  (58)  is  true. 

2.  Prove  that  the  bisectors 
of  two  vertical  angles  are  in 
the  same  straight  line. 

SUGGESTION.  Show  that 
the  sum  of  the  angles  on  one 
side  of  FE  is  equal  to  the 
sum  of  those  on  the  other 
side. 

3.  Prove  that  the  bisectors 
of    two   supplementary   adja- 
cent angles  are  perpendicular 
to  each  other. 

SUGGESTION.  Show  that 
Z  EBF=  i  CBD = I  a  straight  B 

angle. 

4.  If  the  angle  ABD  is  86°  14',  how  many  degrees  are  there  in  each  of 
the  other  angles  formed  at  B  ? 

5.  If  the  angle  ABD  is  two-thirds  of  the  angle  ABC,  how  many 
degrees  are  there  in  each  of  the  other  angles  formed  at  B  ? 

6.  If  the  straight  line  CD  is  the  shortest  line  that   can  be  drawn 
from   C  without  the  line  AB  to  AB,  show  that    CD  is  perpendicular 
to  AB. 

7.  If  a  perpendicular  is  erected  at  the  middle 
point  of  a  line,  any  point  without  the  perpendicu- 
lar is  unequally  distant  from  the  extremities  of 
the  line. 

That  is,  FA  >  FB. 

SUGGESTION.      EA  =  EB ;    add    EF  to  both 
sides  of  this  equation.  A  D  B 

8.  If  any  point  be  taken  within  a  triangle,  show  that  the  sum  of  the 
lines  joining  the  point  to  the  vertices  is  less  than  the  sum  of  the  sides  of 
the  triangle. 


§60.] 


PARALLEL   LINES. 


17 


PARALLEL   LINES. 

59.  DEFINITION.     Two   straight    lines    are    called    Parallel 

when  they  lie  in  the  same  plane,  and  can-       

not   meet  nor  approach   each  other,  how- 
ever far  they  may  be   produced ;    as  AB     „ D 

and  CD. 

60.  AXIOMS.    1.  But  one  straight  line  can  be  drawn  through 
a  given  point  parallel  to  a  given  straight  line. 

2.    Since  parallel  lines  cannot  approach  each  other,  they  are 
everywhere  equally  distant  from  each  other. 


5/8 


If  a  straight  line  EF  cut  two  other  straight  lines  AB  and 
CD,  it  makes  with  those  lines  eight  angles,  to  which  particular 
names  are  given. 

The  angles  1,  4,  6,  7  are  called  Interior  angles. 

The  angles  2,  3,  5,  8  are  called  Exterior  angles. 

The  pairs  of  angles  1  and  7,  4  and  6,  are  called  Alternate- 
interior  angles. 

The  pairs  of  angles  2  and  8,  3  and  5,  are  called  Alternate- 
exterior  angles. 

The  pairs  of  angles  1  and  5,  2  and  6,  4  and  8,  3  and  7,  are 
called  Exterior-interior  angles. 

The  angles  2  and  6,  3  and  7,  4  and  8,  1  and  5,  are  called 
Corresponding  angles. 


C  D 


18  PLANE  GEOMETRY.  [BK.  I. 

PROPOSITION  VIII.     THEOREM. 

61.  Two  straight  lines  perpendicular  to  the  same  straight  line 
are  parallel. 

Let  AB  and  CD  be  perpendicular 
to  CA. 

To  prove  AB  and  CD  are  parallel. 

If  they  are  not  parallel,  they  will 
meet,  and  if  they  meet,  there  will  be 

two  lines  from  this  point  of  meeting  perpendicular  to  the  same 
line,  which  (by  52)  is  impossible. 

Therefore  CD  and  AB,  if  perpendicular  to  AC,  are  parallel. 

EXERCISES. 

1.  Prove  that  two  straight  lines  parallel  to  the  same  straight  line  are 
parallel  to  each  other. 

2.  Prove  that  a  straight  line  perpendicular  to  one  of  two  parallels  is 
also  perpendicular  to  the  other. 

PROPOSITION  IX.     THEOREM. 

62.  If  two  parallel  straight  lines  be  cut  by  a  third  straight  line, 
the  alternate-interior  angles  are  equal. 


D 


Let  AB  and  CD  be  two  parallel  straight  lines  cut  by  the 
line  EF  at  G  and  H. 

To  prove  Z  AGH  =  Z  GHD. 


§02.]  PARALLEL   LINES.  19 

The  lines  AB  and  CD,  being  parallel,  have  the  same  di- 
rection. 

The  lines  EG  and  GH,  being  in  one  and  the  same  straight 
line,  are  similarly  directed. 

That  is,  the  angles  EGB  and  GHD  have  sides  with  the  same 
direction ;  therefore  the  differences  of  their  directions  are 
equal, 

or  (by  30),  Z  EGB  =  Z  GHD. 

But  (by  49),  Z  EGB  =  Z  AGH. 

Therefore  (by  28),  Z  AGH '  =  Z  GHD.  Q.E.D. 

EXERCISES. 

1.  Prove  that  the  alternate-exterior  angles  are  equal, 
or  Z  EGA  -  Z  DHF,  or  Z  EGB  ==  Z  CHF. 

2.  Prove  that  the  sum  of  the  two  interior  angles  on  the  same  side  of 
the  cutting  line,  or  transversal,  is  equal  to  two  right  angles. 

SUGGESTION.     Z  AGH  =  Z  GHD  ;  add  Z  GHC. 

3.  When  two  straight  lines  are  cut  by  a  third  straight  line,  if  the 
exterior-interior  angles  be  equal,  these  two  straight  lines  are  parallel. 


SUGGESTION.     If  AB  is  not  parallel  to  CD,  draw  MN  parallel  to  CD  ; 
then  apply  62  and  28. 


20 


PLANE  GEOMETRY. 


[Bx.  I. 


PROPOSITION  X.     THEOREM. 

• 

63.    Two  angles  whose  sides  are  parallel  each  to  each  are  either 
equal  or  supplementary. 


\ 


F 


Let  AB  be  parallel  to  DH  and  BC  to  KF. 
To  prove  that  the  angle  ABC  is  equal  to  DEF  and  sup- 
plementary to  DEK. 

Let  BC  and  DE  intersect  at  G. 

1.  Since  EG  and  GD  have  the  same  direction,  likewise  EF 
and  6rGY,  the  difference  in  the  directions  of  EG  and  EF  is  the 
same  as  the  difference  in  the  directions  of  GD  and  GrCj  that  is, 


Q.E.D. 


Similarly,  Z  .45(7  = 

(by  28)  Z  .4£<7  =  Z 

2.   Z  G^F  is  the  supplement  of  Z 

.-.  Z  ABC,  which  is  equal  to  Z  GEF.  will  be  the  supplement 

of  Z  (£#jr.  Q.E.D. 


3.    By  (49)  AKEH=  Z 

But  Z  GEF  =  Z  .45(7. 

.-.  by  (28)  Z  KEH=  Z  ^15(7. 

SCHOLIUM.  Two  parallels  are  said  to  be  in  the  same  direction, 
or  in  opposite  directions,  according  as  they  lie  on  the  same  side 
or  on  opposite  sides  of  the  straight  line  joining  their  origins. 


§64.] 


PARALLEL   LINES. 


21 


Thus  AB  and  ED,  and  also  BC  and  EF,  are  in  the  same  direc- 
tion because  they  lie  on  the  same  side  of  BE.  But  BA  and 
EH,  and  also  BC  and  EK,  are  in  opposite  directions. 

63  a.    The  angles  are  equal  when  both  pairs  of  parallel  sides 
extend  in  the  same  direction,  or  in  opposite  directions. 

The  angles  are  supplementary  when  one  pair  of  parallel  sides 
extend  in  the  same  direction 
and  the  other  pair  in  opposite 
directions. 

EXERCISE. 

If  AB  and  CD  bisect  each  other 
at  J7,  show  that  the  straight  lines 
CB  and  AD  are  parallel. 

SUGGESTION.  Apply  CEB  to 
DEA,  and  show  that  Z  A  =  Z  B. 


PROPOSITION  XI.     THEOREM. 

64.    Two  angles  Jiaving  their  sides  perpendicular  each  to  each, 
are  either  equal  or  supplementary. 


Let  DE  be  perpendicular  to  AC,  and  FG  to  AB. 

To  prove  that  the  angle  BAG  is  equal  to  FED  and  supple- 
mentary to  DEG. 

1.  From  A  draw  AH  perpendicular  to  AC,  and  AK  perpen- 
dicular to  LB;  then  AH  is  parallel  to  DE,  and  AKto  FG. 

By  (63)  Z  KAH  =  Z  FED. 


PLANE  GEOMETRY. 


[BK.  I. 


By  construction  the  angles  KAB  and  HAC  are  right  angles, 
and  are  therefore  equal  j  that  is, 

Z  KAB  =  Z  HAC 
or  Z  KAH  +  Z  HAB  =  Z  HAB  +  Z 


subtracting  Z  HAB, 

But 

.-.  (by  28) 

2.    Z  DEG  is  the  supplement  of  Z  FED,  and  is  therefore  the 
supplement  of  the  equal  of  Z  .FZ?/)  or  of  Z  BAG. 


Z  KAH  =^  BAG. 
Z  /&lff  =  Z 

Z  S^IO  =  Z 


TRIANGLES. 

65.  A  Triangle  is  a  plane  figure  bounded  by  three  straight 
lines. 

The  three  straight  lines  which  bound  a  triangle  are  called  its 
Sides.     Thus  AB,  BC,  CA,  are  the  sides  of 
the  triangle  ABC. 

The  angles  of  the  triangle  are  the  angles 
formed  by  the  sides  with  each  other;    as 
BAG,  ABC,  ACS.     The  vertices  of  these 
angles  are  also  called  the   Vertices  of  the  ^~ 
triangle. 

66.  An  Exterior  Angle  of   a  triangle  is   the  angle  formed 
between   any  side  and  the  continuation  of  another  side ;    as 
CAD. 


§75.] 


TRIANGLES. 


23 


The  angles  BAG,  ABC,  BCA,  are  called  Interior  Angles  of  the 
triangle.     When  we  speak  of  the  angles  of  a  triangle,  we  mean 


the  three  interior  angles. 


SCALENE. 


ISOSCELES. 


EQUILATERAL. 


67.  A  Scalene  triangle  is  one  no  two  of  whose  sides  are  equal. 

68.  An  Isosceles  triangle  is  one  two  of  whose  sides  are  equal. 

69.  An  Equilateral  triangle  is  one  three  of  whose  sides  are 
equal. 

70.  The  Base  of  a  triangle  is  the  side  on  which  the  triangle 
is  supposed  to  stand. 

In  an  isosceles  triangle,  the  side  which  is  not  one  of  the 
equal  sides  is  considered  the  base. 


RIGHT. 


OBTUSE. 


ACUTE. 


71.  A  Right  triangle  is  one  which  has  one  of  the  angles  a 
right  angle. 

72.  The  side  opposite  the  right  angle  is  called  the  Hypote- 
nuse. 

73.  An  Obtuse  triangle  is  one  which  has  one  of  the  angles 
an  obtuse  angle. 

74.  An  Acute  triangle  is  one  which  has  all  the  angles  acute. 

75.  An  Equiangular  triangle  is  one  of  which  the  three  angles 
are  equal. 


24  PLANE  GEOMETRY.  [BK.  I. 

76.  When  any  side  has  been  taken  as  the  base,  the  opposite 
angle  is  called  the  Vertical  Angle  and  its  vertex  is  called  the 
vertex  of  the  triangle. 

The  Altitude  of  a  triangle  is  the  perpen- 
dicular drawn  from  the  vertex  to  the  base, 
produced  if  necessary. 

Thus  in  the  triangle  ABC,  BC  is  the  base,    B       D  G 

BAC  the  vertical  angle,  and  AD  the  altitude. 

A  Medial  is  a  line  that  joins  a  vertex  with  the  middle  point 
of  the  opposite  side,  as  BE. 

77.  Since  a  straight  line  is  the  shortest  distance  between 
two  points  (by  6),  it  follows  that  either  side  of  a  triangle  is  less 
than  the  sum  of  the  other  two. 

78.  By  (77)  BC<AB  +  AC. 
Transpose  AB, 

then  BC-AB<AC; 

that  is,  any  side  of  a  triangle  is  greater  than  the  difference  of  the 

other  two  sides. 

PROPOSITION  XII.     THEOREM. 

79.  TJie  sum  of  the  three  angles  of  a  triangle  is  equal  to  two 
right  angles. 

Let  ABC  be  any  triangle. 

To  prove  that  Z  A  +  Z  B  +  /.  BCA  is  equal  to  two  right 
angles. 

From  C  draw  CE  parallel  to  AB. 

By  (63)  Z  EOF  =  Z  A. 

By  (62)  Z  BCE  =  Z.  B. 

But  Z  BCF  =  Z  BCE  +  EOF. 

ZA  +  ZB.  (a) 


§  85.]  TRIANGLES.  25 

Add  the  angle  BCA,  and  we  have 

Z  BCF  +  Z  BOA  =  Z.4-fZ5  +  Z  5O4. 
But,  by  (44),  Z  £<7^  +  Z  BOA  =  2  right  angles. 

/  A  -f  Z  £  +  Z  £(7^1  =  2  right  angles.  Q.E.D. 


80.  COR.  1.     Equation  (a)  when  expressed  in  words  is :  the 
exterior  angle  of  a  triangle  is  equal  to  the  sum  of  the  two  interior 
and  opposite  angles. 

81.  COR.  2.     If  two  angles  of  a  triangle  are  given,  or  merely 
their  sum,  the  third  angle  can  be  found  by  subtracting  this  sum 
from  two  right  angles. 

82.  COR.  3.     If  two  triangles  have  tivo  angles  of  the  one  equal 
to  two  angles  of  the  other,  the  third  angles  are  equal 

83.  COR.  4.     A  triangle  can  have  but  one  right  angle,  or  but 
one  obtuse  angle. 

84.  COR.  5.     In  any  right-angled  triangle  the  tivo  acute  angles 
are  complementary. 

85.  COR.  6.     Each  angle  of  an  equiangular  triangle  is  two- 
thirds  of  a  right  angle. 


26 


PLANE  GEOMETRY. 


[BK.  I. 


EXERCISES. 

1.  If  one  of  the  acute  angles  of  a  right  triangle  is  18°  24'  17",  what  is 
the  value  of  the  other  acute  angle  ? 

2.  If  one  angle  of  a  triangle  is  46°  17',  and  another  is  f  of  a  right 
angle,  what  is  the  value  of  the  other  angle  ? 

3.  If  the  angles  of  a  triangle  are  in  the  proportion  1,  2,  3,  what  is  the 
value  of  each  angle  ? 

4.  How  many  degrees  are  there  in  each  angle  of  an  equiangular  tri- 
angle ? 

5.  If  the  unequal  or  vertical  angle  of   an 
isosceles  triangle  is  46°  18',  what  will  be  the 
value  of  each  of  the  angles  at  the  base  ? 

6.  Show  that  the  sum  of  the  distances  of 
any  point  in  a  triangle  from  the  three  angles 
is  greater  than  half  the  sum  of  the  three  sides 
of  the  triangle. 

DB  + DA  + DC>li(AB+  BC  +  AC}.         B  ^C 

PROPOSITION  XIII.     THEOREM. 

86.  Two  triangles  are  equal  each  to  each,  when  two  sides  and 
the  included  angle  of  the  one  are  equal  respectively  to  two  sides 
and  the  included  angle  of  the  other. 


In  the  triangles  ABC  and  A'B'C',  let  AB  =  A'B',  AC^A'C', 
ZA  =  Z.A. 

To  prove  that  the  triangles  are  equal. 

Place  the  triangle  ABC  upon  the  triangle  A'B'C'  so  that 
the  side  AB  may  fall  upon  A'B',  and  since  AB  =  A'B',  the 
point  B  will  fall  upon  B'. 


§88.]  TRJA\(JLES.  27 

Since  Z.A  =  Z.A',  the  line  AC  will  take  the  direction  of 
A'C',  and  these  lines  being  equal,  the  point  C  will  fall  upon  C'. 

Therefore,  as  the  points  C  and  C",  B  and  B'  are  coincident,  the 
line  joining  B'C1  will  coincide  with  the  line  joining  EC,  or  the 
triangles  will  coincide  throughout,  and  hence  are  equal.  Q.E.D. 

86  a.  Two  right  triangles  are  equal  if  the  two  legs,  or  sides 
including  the  right  angle,  of  the  one  are  equal  to  the  legs  of 
the  other. 

87.  SCHOLIUM.     In  equal  figures,  lines  or  angles  similarly 
situated  are  called  Homologous. 

PROPOSITION  XIV.     THEOREM. 

88.  Two  triangles  are  equal  ivhen  a  side  and  two  adjacent  angles 
of  one  are  equal  respecticely  to  a  side  and  two  adjacent  angles  of 
the  other. 

C' 


B  At 


In  the  triangles  ABC  and  A'B'C',  let  AB  =  A'B',  ZA  = 
ZB=ZB'. 

To  prove  that  the  triangles  are  equal. 

Place  the  triangle  ABC  upon  the  triangle  A'B'C'  so  that 
AB  may  fall  upon  its  equal  A'B'. 

Then,  since  Z  A  =  Z.  A',  the  line  AC  will  take  the  direction 
of  A'C',  and  the  point  C  will  fall  in  the  line  A'C1. 

Since  Z  B  =  Z.  B',  the  line  BC  will  take  the  direction  of 
B'C',  and  the  point  C  will  fall  in  the  line  B'C'. 

.-.  the  point  C,  falling  in  the  lines  A'C'  and  B'C',  it  must 
be  at  the  intersection  of  these  lines,  or  at  the  point  C" ;  that  is, 
the  two  triangles  coincide  throughout  and  are  equal.  Q.E.D. 


28 


PLANE  GEOMETRY. 


[BK.  I. 


89.  COR.  1.    Two  right-angled  triangles  are  equal  when  the 
hypotenuse  and  an  acute  angle  of  the  one  are  equal  respectively 
to  the  hypotenuse  and  an  acute  angle  of  the  other. 

90.  COR.  2.    Two  right-angled  triangles  are  equal  when  a  side 
and  an  acute  angle  of  the  one  are  equal  respectively  to  a  side  and 
homologous  acute  angle  of  the  other. 


PROPOSITION  XV.     THEOREM. 


91.    Two  triangles  are  equal  ivhen  the  three  sides  of  the  one 
are  equal  respectively  to  the  three  sides  of  the  other. 

In  the  triangles  ABC  and  DEF,  AB  =  DE,  AC  =  DF,  and 
BC  =  EF. 

To  prove  that  the  two  triangles 
are  equal. 

Apply  the  triangle  ABC  to 
DEF  so  that  AB  may  coincide 
with  DE  but  the  vertex  C  fall 
on  the  opposite  side  of  DE  from 
F,  that  is,  at  F1,  and  join  FF'. 

By  hypothesis  the  points  D  and 
E  are  equally  distant  from  F  and 
F1 ;  therefore  (by  56)  DH  is  per- 
pendicular to  FF'  at  its  middle  point,  or  the  triangles  DHF, 
DHF',  FHE,  and  F'HE  are  right  triangles. 

The    right    triangles    DHF   and    DHF1    have    DF'  =  DF, 
HF  =  HF',  and  DH  common ;  therefore  (by  86)  they  are  equal, 


B  D 


or 


Z  FDH  =  Z  F'DH  =  Z  BAG. 


This  gives  in  ABC  and  DEF  two  sides  and  the  included 
angle  equal ;  therefore  (by  86)  the  triangles  are  equal.      Q.E.D, 


§92.] 


TRIANGLES. 


29 


PROPOSITION  XVI.     THEOREM. 

92.  Two  right  triangles  are  equal  when  a  side  and  the  hypote- 
nuse of  the  one  are  equal  respectively  to  a  side  and  the  hypotenuse 
of  the  other. 


C     B> 


In  the  right  triangles  ABC  and  A'B'C',  let  AB  =  A'B>,  and 
AC=A'C'. 

To  prove  that  the  triangles  are  equal. 

Apply  the  triangle  ABC  to  A'B'C',  so  that  BC  will  take  the 
direction  of  B'C'. 

Then,  since  Z.  B  =  Z  B',  both  being  right  angles,  the  side  BA 
will  take  the  direction  of  B'A,  and  since  BA  —  B'A  the  point 
A  will  fall  on  A'. 

Since  A C=  AC'  (by  53),  they  will  cut  off  equal  distances 
from  the  foot  of  the  perpendicular ;  that  is, 

BC  =  B'C'. 

Therefore  the  triangles  ABC  and  A'B'C'  having  three  sides 
equal  are  (by  91)  equal  in  all  their  parts.  Q.E.D. 


EXERCISES. 

1.  Prove  that  in  any  obtuse-angled  triangle  the  sum  of  the  acute  angles 
is  less  than  a  right  angle. 

2.  Prove  that  in  any  acute-angled  triangle  the  sum  of  any  two  acute 
angles  is  greater  than  a  right  angle. 


30  PLANE  GEOMETRY.  [Bit.  I. 


PROPOSITION  XVII.     THEOREM. 

93.  In  an  isosceles  triangle  the  angles  opposite  the  equal  sides 
are  equal. 

Let  ABC  be  an  isosceles  triangle  in  which 
AC  and  BC  are  the  equal  sides. 

To  prove  that  /.A  =  Z.  B. 

Draw  CD  perpendicular  to  AB. 

Then  the  triangles  ADC  and  BDC  having 
the  side  CD  common  and  the  hypotenuses 
equal,  are  (by  92)  equal  in  all  their  parts,  and 
/-  A  =  Z  B.  Q.E.D. 

94.  COR.  1.     The  equality  of  the  triangles  ADC  and  BDC 
also  gives  AD  —  DB. 

Hence  the  straight  line  ivhicli  bisects  the  vertical  angle  of  an 
isosceles  triangle  bisects  the  base  at  right  angles. 

And,  in  general,  if  a  straight  line  is  drawn  so  as  to  satisfy 
any  two  of  the  following  conditions, 

1.  Passing  through  the  vertex, 

2.  Bisecting  the  vertical  angle, 

3.  Bisecting  the  base, 

4.  Perpendicular  to  the  base, 

it  will  also  satisfy  the  remaining  conditions. 

EXERCISES. 

1.  Show  conversely,  if  two  angles  of  a  tri- 
angle are  equal,  the  sides  opposite  are  equal 
and  the  triangle  is  isosceles. 

2.  Show  that  if  the  equal  sides  of  an  isos- 
celes triangle  be  produced,  the  angles  formed         

with  the  base  by  the  sides  produced  are  equal.     /  \ 


§95.] 


TRIANGLES. 


3.  Show  that  if  the  perpendicular  from  the 
vertex  to  the  base  of  a  triangle  bisects  the  base, 
the  triangle  is  isosceles. 

4.  How  many  degrees  are  there  in  the  ex- 
terior angle  at  each  vertex  of  an  equiangular 
triangle  ? 

5.  Show  that  the   bisectors   of    the   equal 
angles  of   an  isosceles  triangle  form  with  the 
base  another  isosceles  triangle ;  that  is,  DBG 
is  isosceles. 

6.  What  are  the  relative  values  of  the  verti- 
cal angles  D  and  A  in  the  above  ? 

7.  Show  that  a  straight  line  parallel  to  the 
base  of  an  isosceles  triangle  makes  equal  angles 
with  its  sides,  or  Z  D  =  Z  E. 


PROPOSITION  XVIII.     THEOREM. 

95.    Of  tivo  sides  of  a  triangle,  that  is  the  greater  which  is  oppo- 
site the  greater  angle. 


In  the  triangle  ABC  let  angle  ACB  be  greater  than  angle  B. 
To  prove  that  AB  >  AC. 
From  C  draw  CE,  making  Z  ECB  =  Z  B. 
Then  the  triangle  BEG  is  isosceles  and  the  side  EB  =  EC. 
Add  AE,  AE  +  EB  =  AE  +  EC, 

or  AB  =  AE  +  EC. 

But  (by  77)  AE  +  EC  >  AC. 

.:  AB,  which  is  equal  to  AE  -f  EC,  is  greater  than  AC. 

Q.E.D. 


32  PLANE  GEOMETRY.  [BK.  I. 

EXERCISE. 

1.    Show  that  of  two  angles  of  a  triangle,  that  is  the  greater  which  is 
opposite  the  greater  side. 

PROPOSITION  XIX.     THEOREM. 

96.    The  three  bisectors  of  the  angles  of  a  triangle  meet  in  a 
point.  n 

\ 

.K 

i      \ 

H, 


P 

Let  AO  and  CO  be  the  bisectors  of  the  angles  A  and  C 
of  the  triangle  ABC. 

To  prove  that  the  bisectors  meet  in  a  point. 

Suppose  AO  and  CO  meet  at  0,  and  join  BO. 

Let  fall  the  perpendiculars,  OP,  OH,  and  OK,  forming  the 
right  triangles  AOP,  AOH,  COP,  and  COK. 

The  triangles  AOP  and  AOH  are  equal  (by  89),  having 
the  hypotenuse  AO  common,  and  Z  OAP=  Z  OAH;  therefore 

OP  =  OH. 

For  the  same  reason,  the  triangles  OPC  and  OKC  are  equal, 
or  OP=  OK, 

or  (by  28)  OH=  OK. 

The  two  right  triangles  BOH  and  BOK  have  BO  common, 
and  OH=  OK]  therefore  (by  92)  they  are  equal;  that  is, 

Z  HBO  =  Z  KBO, 
or  BO  is  a  bisector  of  Z  B. 

.-.  the  three  lines  meeting  in  0  are  bisectors  of  the  angles. 

97.  COR.  Since  OP  =  OK=  OH,  it  is  shown  that  the  bisec- 
tors of  angles  are  equally  distant  from  their  sides. 


§99.J 


Q  UADRILA  TERALS. 


33 


EXERCISES- 

1.  Show  that  the  three  perpendiculars  erected  at  the  middle  points 
of  the  three  sides  of  a  triangle  meet  in  a  point,  and  this  point  is  equally 
distant  from  the  vertices. 

SUG.     See  53. 


2.  If  an  exterior  angle  is  formed  at 
each  vertex  of  a  triangle,  their  sum  will 
be  equal  to  four  right  angles. 


3.  Show  that  the  perpendiculars  from 
the  vertices  of  a  triangle  to  the  opposite 
sides  meet  in  a  point. 


4.  Show  that  every  point  unequally 
distant  from  the  sides  of  an  angle  lies 
outside  of  the  bisector  of  that  angle. 


QUADRILATERALS. 
DEFINITIONS. 

98.  A  Quadrilateral  is  a  plane  figure 
bounded  by  four  straight  lines ;  as  ABCD. 

99.  The  bounding  lines  are  called  the 
Sides  of  the  quadrilateral,  and  their  points 
of  intersection  are  called  its  Vertices 


34 


PLANE   GEOMETRY, 


100.  The  Angles  of  the  quadrilateral  are  the  interior  angles 
formed  by  the  sides  with  each  other. 

101.  A  Diagonal  of  a  quadrilateral  is  a  straight  line  joining 
two  vertices  not  adjacent;  as  AC. 


102.   Quadrilaterals  are  divided  into  classes  as  follows: 

1st.  The  Trapezium  (A),  which  has  no  two 
of  its  sides  parallel. 

2d.  The  Trapezoid  (IB),  which  has  two  sides 
parallel.  The  parallel  sides  are  called  the 
Bases,  and  the  perpendicular  distance  between 
them  the  Altitude  of  the  trapezoid. 

3d.  The  Parallelogram  ( C),  which  is  bounded 
by  two  pairs  of  parallel  sides. 


B 


103.   The  side  upon  which  a  parallelogram 
is  supposed  to  stand  and  the  opposite  side  are 
called  its  lower  and  upper  Bases.     The  perpendicular  distance 
between  the  bases  is  the  Altitude. 


104.  Parallelograms  are  divided  into  species 
as  follows : 

The  Rhomboid  (A),  whose  adjacent  sides 
are  not  equal  and  whose  angles  are  not  right 
angles. 

The  Rhombus  (B)}  whose  sides  are  all  equal. 

The  Rectangle  (C),  whose  angles  are  all  right 
angles. 

The  Square  (Z>),  whose  sides  are  all  equal 
and  whose  angles  are  all  equal. 

105.  The  square  is  at  once  equilateral  and 
equiangular. 


§  108.]  QUADRILATERALS.  35 

PROPOSITION  XX.     THEOREM. 

106.   In  a  parallelogram  the  opposite  sides  are  equal,  and  the 
opposite  angles  are  equal. 

C 


Let  the  figure  ABCE  be  a  parallelogram. 
To  prove  that  AB  =  CE,  and  BC  =  AE,  and  ^B  =  Z.E, 
Z  .1  =  Z  C. 

Draw  the  diagonal  AC. 

Since  AB  and  CE  are  parallel  and  AC  cuts  them, 

(by  62),  Z BAG  =  Z  ACE. 

Since  AE  and  BC  are  parallel  and  AC  cuts  them, 
(by  62),  Z  ACB  =  Z  CAE. 

Then  the  triangles  ABC  and  ACE  have  the  side  AC  com- 
mon, and  the  two  adjacent  angles  equal ;  they  are,  therefore 
(by  88),  equal  in  all  their  parts, 

or  AB=CE,  BC=AE,  and  ZB  =  ZE. 

Likewise,  since  Z  BAG  =  Z  ACE. 
and  *  Z  CAE  =  Z  ACB. 

By  addition,  Z  BAE  =  Z  BCE, 

or  Z  .4  =  Z  (7.  Q.E.D. 

107.  COR.  1.  A  diagonal  of  a  parallelogram  divides  it  into 
two  equal  triangles. 

108.  COR.  2.     Parallel   lines  included   between  parallels   are 
equal. 


36 


PLANE   GEOMETRY. 


[BK.  I. 


D 


c 


EXERCISES. 

1.  Show  conversely,  that  if  the  opposite  sides  of  a  quadrilateral  are 
equal,  the  figure  is  a  parallelogram. 

2.  If  two  sides  of  a  quadrilateral  are  equal     A 
and  parallel,  the  figure  is  a  parallelogram. 

3.  If   one  angle  of  a  parallelogram  is  a 
right  angle,  the  figure  is  a  rectangle. 

4.  If    two   parallels  are   cut  by   a  third 
straight  line,  the  bisectors  of  the  four  interior 
angles  form  a  rectangle.     (See  58,  Ex.  3.) 

5.  If  CE  is  drawn  parallel  to  BD,  meeting 
AD  produced,  show  that  BCED  is  a  parallel- 
ogram and  equal  to  the  parallelogram  ABCD. 

PROPOSITION  XXI.     THEOREM. 

109.    The  diagonals  of  a  parallelogram  bisect  each  other. 

B  C 


A  E 

Let  the  figure  ABCE  be  a  parallelogram,  and  let  the 
diagonals  AC  and  BE  cut  each  other  at  0. 

To  prove  that     AO  =  OC  and  BO  =  OE. 

In  the  triangles  BOC  and  AOE,  BC  =  AE  (by  106),  Z.  BOO 
=  Z.  OAE,  and  /.  OBC  =  Z  OEA  (by  62) ;  the  triangles  are 
therefore  equal  (by  88)  in  all  their  parts, 

or  BO  =  OE  and  AO  =  OC.  Q.E.D. 

EXERCISES. 

1.  Show  conversely,  if  the  diagonals  of  a  quadrilateral  bisect  each 
other,  the  figure  is  a  parallelogram. 


§111.]  QUADRILATERALS.  37 

2.  Show  that  the  diagonals  of  a  rhombus  bisect  each  other  at  right 
angles. 

3.  Show  that  the  diagonals  of  a  rectangle  are  equal. 

4.  Show  that  two  parallelograms  are  equal  when  two  adjacent  sides 
and  the  included  angle  of  the  one  are  equal  to  the  two  adjacent  sides  and 
the  included  angle  of  the  other. 

PROPOSITION  XXII.     THEOREM. 

110.    If  three  or  more  parallels  intercept  equal  lengths  on  any 
transversal,  they  intercept  equal  lengths  on  every  transversal. 

Let  AE,  BF,  CG,  and  DHbe  parallels, 
and  MN  and  OP  any  two  transversals. 

To  prove  that  if  AB  =  BC  =  CD, 
EF=FG  =  GH. 


Draw  EK,  FL,  and   GS  parallel  to 


F 


K 


G 


H 


MN. 

Then,  since  EK  and  AB  are  parallels 
included  between  parallels,  they  are 
equal  (by  108).  Likewise,  FL  =  BC, 
and  GS  =  CD.  N  ** 

But  AB  =  BC=  CD  by  hypothesis, 

then  (by  28)  EK=  FL  =  GS. 

In  the  triangles  EKF,  FLG,  and  GSH  the  angles  KEF, 
LFG,  and  SGH  are  equal  (by  63);  also  the  angles  EFK, 
FGL,  and  GHS  are  equal  (by  63). 

Therefore  these  triangles  are  (by  88)  equal  in  all  their  parts, 

or  EF=FG=  GH.  Q.E.D. 

111.  COR.  From  the  equality  of  the  triangles  EKF,  FLG, 
and  GSH,  KF=LG  =  SH. 

HD--DS  =  SH,  but  (by  108)  CG  =  DS;  therefore  HD 
-CG  =  SH-,  likewise,  CG  -  BF =  LG,  and  BF-  AE  =  KF. 

Therefore  the  intercepted  part  of  each  parallel  will  differ  in 
length  from  the  next  intercept  by  the  same  amount. 


38  PLANE   GEOMETRY.  [BK.  I. 


PROPOSITION  XXIII.     THEOREM. 

112.  The  straight  line  drawn  through  the  middle  point  of  a 
side  of  a  triangle  parallel  to  the  base  bisects  the  remaining  side, 
and  is  equal  to  half  the  base. 

In  the  triangle  ABC  let  E  be  the  middle 
point  of  AC  and  DE  parallel  to  BC. 

To  prove  that  D  is  the  middle  point  of  AB 
and  that  DE  =  ±BC. 


Through  A  draw  a  line  parallel  to  DE,  and 


. 
it  will  be  parallel  to  BC. 

Then  AB  and  AC  are  transversals  cutting  parallel  lines; 
therefore  (by  110),  when  AE  =  EC,  AD  =  DB.  Q.E.D. 

Likewise  (by  111),  BC  -  DE  =  DE  -  AA  '  =  DE  -  0  =  DE. 
Transpose  DE,  then     BC=2  DE, 
or  DE  =  ±  BC.  Q.E.D. 


PROPOSITION  XXIV.     THEOREM. 

113.  The  line  drawn  parallel  to  the  bases  through  the  middle 
point  of  one  of  the  non-parallel  sides  of  a  trapezoid  bisects  the 
opposite  side,  and  is  equal  to  half  of  the  parallel  sides. 


7~) 


r  "V* 

I_  _  \ 


Let  ABCD  be  a  trapezoid,  GH  a  line 
drawn  from  G,  the  middle  point  of  AD 
parallel  to  AB. 

To  prove  that  GH=±(AB+DC),     AL  _  \B 
and  HB  =  CH. 

Since  DA  and  CB  are  transversals,  and  DC,  GH,  and  AB 
parallels  (by  110),  when  DG  =  GA,  CH=  HB.  '  Q.E.D. 

Also  (by  111)  AB-GH=GH-  CD,  or  AB  +  CD  =  2  GH] 


§114.]  QUADEILATERALS.  39 


PROPOSITION   XXV.     THEOREM. 

114.  The  three  medial  lines  of  a  triangle  meet  in  a  point 
which  is  two-thirds  of  the  icay  from  each  angle  to  the  middle  of 
the  opposite  side. 

Let  ABC  be  a  triangle ;  P,  Q,  R,  the  middle  points  of  its 
respective  sides;  BR,  AQ,  two  medial  lines  of  the  triangle; 
0,  their  point  of   inter- 
section. 

To  prove  that  the  third 
medial  line  CP  passes 
through  0,  and  that 
CO  =  ICP,AO  =  %AQ, 
and  BO  =  f  BR. 

Bisect  AO  in  M,  and 
BO  in  N;  join  RM  and  A 
Q.Yand  0(7. 

In  the  triangle  AOC,  M  is  the  middle  point  of  AO  by  con- 
struction, and  R  the  middle  point  of  AC  by  hypothesis  ;  there- 
fore (by  112)  RM  is  parallel  to  CO  and  equal  to  one-half  of 
CO. 

In  the  triangle  BOC,  for  the  same  reason,  NQ  =  ±  CO  and  is 
parallel  to  CO. 

Therefore  (by  28)  RM  is  equal  to  and  parallel  with  QN. 

In  the  triangle  ACB  (by  112),  RQ  =  1  AB  and  is  parallel 
to  it. 

In  the  triangle  AOB  (by  112),  MN=  i  AB  and  is  parallel 
to  it. 

Therefore  RQ  =  MN  said  is  parallel  to  it. 

Hence  the  figure  RMNQ  is  (by  102)  a  parallelogram. 

Since  RMNQ  is  a  parallelogram,  (by  109)  OR=  ON,  and 
MO  =  OQ. 

But  by  construction  OM=  AM-,  therefore  the  three  parts 
AM,  MO,  and  OQ  into  which  AQ  is  divided  are  equal. 


40 


PLANE  GEOMETRY. 


[BK.  I. 


Therefore  AO,  which  contains  two  of  these  parts,  is  two- 
thirds  of  the  whole,  or  AO  =  f  AQ,  and  likewise  BO  =  f  BR. 

Q.E.D. 

By  taking  CP  and  BR  as  medial  lines  intersecting  at  0,  and 
joining  S,  the  middle  point  of  GO,  with  N,  and  with  jR,  and 
drawing  RP  and  JVP,  it  can  be  shown  in  the  same  manner  that 
OC  =  f  CP,  and  that  0  is  a  point  on  all  three  medial  lines. 

EXERCISES. 

1.  The  bisectors  of  the  interior  angles  of  a  parallelogram  form  a 
rectangle. 

2.  If  the  non-parallel  sides  of  a  trapezoid  are 
equal,  the  angles  which  they  make  with  the  bases 
are  equal. 

3.  If  from  any  point  in  the  base  of  an  isosceles 
triangle  parallels  to  the  equal  sides  are  drawn,  the 
perimeter  of    the   parallelogram  thus  formed  is 
equal  to  the  sum  of  the  equal  sides  of  the  triangle. 

SUGGESTION.     See  112. 

115.  A  Polygon  is  a  plane  figure  bounded  by  straight  lines ; 
as  ABODE. 

The  straight  lines  are  called  the  Sides  of  the  polygon ;  and 
their  sum  is  called  the  Perimeter  of  the 
polygon. 

The  Angles  of  the  polygon  are  the  angles 
formed  by  the  adjacent  sides  with  each 
other ;  and  the  vertices  of  these  angles 
are  also  called  the  Vertices  of  the  polygon. 

116.  The  angles  of  the  polygon  within       D 
the  polygon  and  included  between  its  sides 

are  called  Interior  Angles. 

An  Exterior  Angle  of  a  polygon  is  an  angle  between  any  side 
and  the  continuation  of  an  adjacent  side. 

A  Diagonal  is  a  line  joining  any  two  vertices  that  are  not 
adjacent,  as  AD. 


§  120.] 


POLYGONS. 


41 


117.    Polygons  are  named  from  the  number  of  their  sides, 
as  follows : 


No.  OF 
SIDES. 

DESIGNATION. 

No.  OF 

SIDES. 

DESIGNATION. 

3 

Triangle. 

8 

Octagon. 

4 

Quadrilateral. 

9 

Enneagon. 

5 

Pentagon. 

10 

Decagon. 

6 

Hexagon. 

11 

Hendecagon. 

7 

Heptagon. 

12 

Dodecagon,  etc. 

118.   An  Equilateral  polygon  is  one  all  of  whose  sides  are 
equal. 

An   Equiangular   polygon  is  one  all  of 
whose  angles  are  equal. 


119.  A  polygon  is  called  Convex  when 
each  of  its  angles  is  less  than  a  straight 
angle ;  as  ABCDE. 

It  is  evident  that  in  such  a  polygon  no 
side,  if  produced,  can  enter  the  space  enclosed  by  the  perimeter. 

120.  A  polygon  is  called  Concave  when  at  least  one  of  its 
angles  is  greater  than  a  straight   angle ;    as 

FGHIKj  in  which  the  interior  angle  whose 
vertex  is  H  is  greater  than  a  straight  angle. 

Such  an  angle  is  called  Reentrant. 

It  is  evident  that  in  such  a  polygon  at  least 
two  sides,  if  produced,  will  enter  the  space 
enclosed  by  the  perimeter. 

All  polygons  treated  hereafter  will  be  understood  to  be  con- 
vex, unless  the  contrary  is  stated. 


42 


PLANE   GEOMETRY. 


[BK.  I. 


121.  Two  polygons,  ABODE,  AB'C'D'E',  are  equal  when 
they  can  be  divided  by  di- 
agonals into  the  same  num- 
ber of  triangles,  equal  each 

to  each,  and  similarly  ar- 
ranged; for  the  polygons 
can  evidently  be  superposed, 
one  upon  the  other,  so  as  to 
coincide. 

122.  Two    polygons    are    mutually    equiangular   when    the 
angles  of  the  one  are  re- 
spectively   equal    to    the 

angles  of  the  other,  taken 

in    the    same    order;    as 

ABCD,      A'B'C'D',      in 

which    A  =  A',    B  =  B', 

etc.    The  equal  angles  are 

called  Homologous  Angles;  the  sides  containing  equal  angles, 

and  similarly  placed,  are  Homologous  Sides;  thus  A  and  A'  are 

homologous  angles,  AB  and  A'B'  are  homologous  sides,  etc. 

Two  polygons  are  mutually  equilateral  when  the  sides  of  the 
one  are  respectively  equal  to 
the  sides  of  the  other,  taken 
in  the  same  order ;  as  MNPQ, 
M'N'P'Q',  in  which  MN 
=  M'N',  NP  =  N'P',  etc. 
The  equal  sides  are  homolo- 
gous ;  and  angles  contained  by  equal  sides  similarly  placed, 
are  homologous ;  thus  MN  and  M'N'  are  homologous  sides ; 
M  and  M'  are  homologous  angles. 

Two  mutually  equiangular  polygons  are  not  necessarily  also 
mutually  equilateral.  Nor  are  two  mutually  equilateral  poly- 
gons necessarily  also  mutually  equiangular,  except  in  the  case 
of  triangles  (91). 


§124.]  POLYGONS.  43 

If  two  polygons  are  mutually  equilateral  and  also  mutually 
equiangular,  they  are  equal ;  for  they  can  evidently  be  super- 
posed, one  upon  the  other,  so  as  to  coincide. 


PROPOSITION   XXVI.     THEOREM. 

123.  The  sum  of  the  interior  angles  of  a  polygon  is  equal  to 
tivo  right  angles  taken  as  many  times  as  the  polygon  has  sides  less 
two. 

F  E 


B 


Let  ABCDEF  be"  a  polygon,  and  AE,  AD,  and  AC  diagonals. 

These  diagonals  divide  the  polygon  into  triangles. 

Since  the  first  and  last  triangles  involve  two  sides  of  the 
polygon,  while  each  other  triangle  only  involves  one  side  of 
the  polygon,  there  will  always  be  two  triangles  less  than  the 
number  of  sides  in  the  polygon. 

The  sum  of  the  angles  of  the  polygon  will  be  equal  to  the 
sum  of  the  angles  of  the  triangles,  but  (by  79)  the  angles  of 
each  triangle  are  equal  to  two  right  angles ;  therefore,  since  the 
number  of  triangles  is  two  less  than  the  number  of  sides  in 
the  polygon,  the  angles  of  the  polygon  are  equal  to  two  right 
angles  taken  as  many  times,  less  two,  as  the  figure  has  sides. 

Q.E.D. 

124.  COR.  The  sum  of  the  angles  of  a  quadrilateral  is  equal 
to  four  right  angles ;  of  a  pentagon,  six  right  angles;  of  a  hexa- 
gon, eight  right  angles;  etc. 


44  PLANE   GEOMETRY.  [BK.  I. 

125.  SCHOLIUM.    If  R  denotes  a  right  angle,  and  n  the  num- 
ber of  sides  of  the  polygon,  the  sum  of  its  angles  is  expressed 
by  2  It  x  (71  -  2),  or  2  nR  -  4  R. 

That  is,  the  sum  of  the  angles  of  a  polygon  is  equal  to  twice  as 
many  right  angles  as  the  figure  has  sides,  less  four  right  angles. 

PROPOSITION   XXVII.     THEOREM. 

126.  The  exterior  angles  of  a  polygon,  made  by  producing  each 
of  its  sides  in  succession,  are  together  equal  to  four  right  angles. 


Let  the  figure  ABODE  be  a  polygon,  having  its  sides  pro- 
duced in  succession. 

To  prove  that  the  sum  of  the  angles,  a,  b,  c,  d,  and  e  is 
equal  to  four  right  angles. 

The  sum  of  each  exterior  and  its  corresponding  interior 
angle  (by  44)  is  equal  to  two  right  angles. 

That  is,  the  sum  of  the  interior  and  exterior  angles  is  equal 
to  twice  as  many  right  angles  as  the  figure  has  sides. 

But  by  (125)  the  interior  angles  are  equal  to  twice  as  many 
right  angles  as  the  figure  has  sides,  less  four  right  angles. 

Therefore  the  exterior  angles  alone  are  equal  to  four  right 
angles.  Q.E.D. 

EXERCISES. 

1.  If  one  side  of  a  regular  hexagon  is  produced,  show  that  the  exterior 
angle  is  equal  to  the  angle  of  an  equilateral  triangle. 


§  126.]  POLYGONS.  45 

2.  The  exterior  angle  of  a  regular  polygon  is  18°  ;  find  the  number  of 
sides  in  the  polygon. 

3.  The  interior  angle  of  a  regular  polygon  is  five-thirds  of  a  right 
angle  ;  find  the  number  of  sides  in  the  polygon. 

4.  How  many  degrees  are  there  in  each  angle  of  a  regular  pentagon  ? 
Of  a  regular  hexagon  ?     Of  a  regular  dodecagon  ? 

5.  If  two  angles  of  a  quadrilateral  are  supplementary,  the  other  two 
angles  are  supplementary. 

6.  If  a  diagonal  of  a  quadrilateral  bisects  twro  of  its  angles,  it  is  per- 
pendicular to  the  other  diagonal. 


BOOK  II. 

THE   CIRCLE. 


DEFINITIONS. 

127.  A  Circle  is  a  plane  figure  bounded  by  a  curve,  all 
points  of  which  are  equally  distant  from  a  point  within  called 
the  Centre. 

The  curve  which  bounds  the  circle  is 
called  the  Circumference,  and  any  portion 
of  it  is  called  an  Arc.  A 


128.  A  Chord  is  a  straight  line  which 
joins  any  two  points  on  the  circumference, 
as  BC. 

When  a  chord  passes  through  the  centre,  it  has  its  greatest 
length,  and  is  called  the  Diameter. 

129.  A  Radius  is  a  straight  line  drawn  from  the  centre  to 
the  circumference,  and  since,  by  definition,  this  distance  is  the 
same  for  the  same  circle,  all  radii  are  equal ;  and  each  radius 
is  one-half  of  the  diameter. 

130.  An  arc  equal  to  one-half  the  circumference  is  called  a 
Semi-circumference,  and  an  arc  equal  to  one-fourth  of  the  cir- 
cumference is  called  a  Quadrant. 

131.  Two  circles  are  Equal  when  they  have  equal  radii,  for 
they  can  evidently  be  applied  one  to  the  other  so  as  to  coincide 

throughout. 

46 


BK.  I.  §  140.] 


THE   CIRCLE. 


47 


132.  Two  circles  are  Concentric  when  they  have  the  same 
centre. 

133.  POSTULATE  :  the  circumference  of  a  circle  can  be  de- 
scribed about  any  point  as  a  centre  and  with  any  distance  for 
a  radius. 

134.  A  Segment  of  a  circle  is  a  portion  of  a  circle  enclosed 
by  an  arc  and  its  chord,  as  AMB,  Fig.  1. 

135.  A  Semicircle  is  a  segment  equal  to  one-half  the  circle, 
as  ADC,  Fig.  1. 

136.  A  Sector  of  a  circle  is  a  portion  of  the  circle  enclosed 
by  two  radii  and  the  arc  which  they  intercept,  as  ACB,  Fig.  L*. 

137.  A  Tangent  is  a  straight  line  which  touches  the  circum- 
ference, but  does  not  intercept  it,  however  far  produced.     The 
point  in  which  the  tangent  touches  the  circumference  is  called 
the  Point  of  Contact,  or  Point  of  Tangency. 

138.  Two  Circumferences  are  tangent  to   each  other  when 
they  are  tangent  to  a  straight  line  at  the  same  point. 

139.  A  Secant  is  a  straight  line  which  intersects  the  circum- 
ference in  two  points,  as  AD,  Fig  3. 


D 


140.    A  straight  line  is  Inscribed  in  a  circle  when  its  ex- 
tremities lie  in  the  circumference  of  the  circle,  as  AB,  Fig.  1. 
An  angle  is  inscribed  in  a  circle  when  its  vertex  is  in  the 


48 


PLANE  GEOMETRY. 


[BK.  II. 


circumference,  and  its  sides  are  chords  of  that  circumference, 
as  Z  ABC,  Fig.  1. 

A  polygon  is  inscribed  in  a  circle  when  its  sides  are  chords 
of  the  circle,  as  ABC,  Fig.  1. 

A  circle  is  inscribed  in  a  polygon  when  the  circumference 
touches  the  sides  of  the  polygon  but  does  not  intersect  them, 
as  in  Fig.  4. 


D 


D 


141.  A  polygon  is  Circumscribed  about  a  circle  when  all  the 
sides  of  the  polygon  are  tangents  to  the  circle,  as  in  Fig.  4. 

A  circle  is  circumscribed  about  a  polygon  when  the  circum- 
ference passes  through  all  the  vertices  of  the  polygon,  as 
in  Fig.  1. 

142.  Every  diameter  bisects  the  circle  and  its  circumference. 
For  if  we  fold  over  the  segment  AMB  on 

AB  as  an  axis  until  it  comes  into  the  plane 
of  APB,  the  arc  AMB  will  coincide  with 
the  arc  APB ;  because  every  point  in  each 
is  equally  distant  from  the  centre  0. 

143.  A  straight  line  cannot  meet  the  cir- 
cumference  of  a   circle   in   more    than    ttvo 

points.  For  if  it  could  meet  it  in  three  points,  these  three 
points  would  be  equally  distant  from  the  centre  (127).  There 
would  then  be  three  equal  straight  lines  drawn  from  the  same 
point  to  the  same  straight  line,  which  is  impossible  (58  a). 


§  145.] 


THE  CIRCLE. 


49 


PROPOSITION  I.     THEOREM. 

144.    In   equal   circles,  or  in  the  same  circle,  equal  arcs  are 
intercepted  by  equal  central  angles  and  have  equal  chords. 


M  Mr 

Let   ABM  and  A'B'M'  be    two    equal    circles,   in   which 

To  prove  that  the  arc  AB  =  arc  A'B'  and  the  chord  AB 
=  chord  A'B'. 

1.  Place  the  circle  ABM  upon  A'B'M'  so  that  their  centres 
may  coincide,  and  A  fall  upon  A' ;  then,  since  they  are  equal 
circles,  they  will  coincide  throughout. 

Since  Z.  C  =  Z.  C',  the  radius  CB  will  take  the  direction  of 
OB',  and,  being  radii  of  equal  circles,  B  will  fall  upon  B'. 

Therefore  the  arc  AB  will  coincide  with  arc  A'B'  and  be 
equal  to  it.  Q.E.D. 

2.  The  two  triangles  ACB  and  A'C'B'  have  AC  =  AC1  and 
BC=B'C',  being  radii  of  equal  circles   (by  131),   and   Z.  C 
=  /.  C'  by  hypothesis. 

Therefore  (by  86)  the  two  triangles  are  equal  in  all  their 
parts,  or  AB  =  A'B'.  Q.E.D. 

145.  COR.  1.  Another  form  of  statement  is :  In  equal  circles, 
or  in  the  same  circle,  equal  central  angles  intercept  equal  arcs  on 
the  circumference. 

COR.  2.  Also  the  converse :  In  equal  circles,  or  in  the  same 
circle,  equal  chords  subtend  equal  arcs  and  equal  angles  at  the 
centre. 


50 


PLANE  GEOMETRY. 


[BK.  II. 


PROPOSITION  II.     THEOREM. 

146.    The  diameter  perpendicular  to  a  chord  bisects  the  chord 
and  its  subtended  arcs. 


B 


In  the  circle  ADB,  let  the  diameter  CD  be  perpendicular  to 
the  chord  AB. 

To  prove  that  DC  bisects  AB  and  its  subtended  arcs. 

Let  0  be  the  centre  of  the  circle,  and  join  OA  and  OB. 

Then  since  OA=  OB,  the  triangle  OAB  is  isosceles;  and 
the  line  CD,  passing  through  the  vertex  perpendicular  to  the 
base,  bisects  the  base  and  also  the  vertical  angle  (94). 

Hence  Z  AOC  =  Z.  BOG,  and  arc  AC  =  arc  BC  (144). 

Subtracting  the  equal  arcs  AC  and  BC  from  the  semicircum- 
f erences  CAD  and  CBD,  we  have  arc  AD  =  arc  BD. 

Therefore  the  diameter  bisects  the  chord  AB  and  its  sub- 
tended arcs  ACB  and  ADB. 

147.  COR.  The  perpendicular  erected  at  the  middle  point  of  a 
chord  passes  through  the  centre  of  the  circle. 

And  in  general,  if  a  straight  line  is  drawn  so  as  to  satisfy 
any  two  of  the  following  conditions : 

1.  Passing  through  the  centre, 

2.  Perpendicular  to  the  chord, 

3.  Bisecting  the  chord, 

4.  Bisecting  the  less  subtended  arc, 

5.  Bisecting  the  greater  subtended  arc, 

it  will  also  satisfy  the  remaining  conditions. 


§  149.]  THE   CIRCLE.  51 


PROPOSITION  III.     THEOREM. 

148.  Li  the  same  circle,  or  equal  circles,  equal  chords  are 
equally  distant  from  the  centre;  and  of  two  unequal  chords  the 
less  is  at  the  greater  distance  from  the  centre. 


In  the  circle  ABEC  let  the  chord  AB  equal  the  chord  CF, 
and  the  chord  CE  be  less  than  the  chord  CF.  Let  OP,  OH, 
and  OK  be  perpendiculars  drawn  to  these  chords  from  the 
centre  0. 

To  prove  that  OP=  OH,  and  that  OK>  OP  or  OH. 
Join  OA  and  OC. 

1.  The  right  triangles  OAP  and  OCH  have  by  hypothesis 
AP  =  CH,  and  OA=OC  being  radii,  therefore   (by  92)  the 
triangles  are  equal  in.  all  their  parts,  or  OP  =  OH.  Q.E.D. 

2.  Since  Om  is  an  oblique  line,  and  OH  a  perpendicular, 

Om  >  OH,  but  OK>  Om, 
therefore  OK>  OH  or  its  equal  OP.  Q.E.D. 

149.  COR.  The  conclusion  is  reached  that  the  nearer  the  centre 
the  greater  the  chord,  therefore  the  greatest  chord,  is  at  no  dis- 
tance from  the  centre,  or  passes  through  the  centre,  that  is  the 
diameter  (128). 


52 


PLANE  GEOMETRY. 


[BK.  II. 


EXERCISE. 

From  a  point  within  the  circle,  other  than 
the  centre,  not  more  than  two  equal  straight 
lines  can  be  drawn  to  the  circumference. 
(See  147.) 


PROPOSITION  IV.     THEOREM. 

150.   A  straight  line  perpendicular  to  a  radius  at  its  extremity 
is  a  tangent  to  the  circumference. 


Let  BC  be  perpendicular  to  the  radius  OA  at  its  extremity  A. 
To  prove  that  BC  is  a  tangent  to  the  circumference. 

Since  OA  is  the  shortest  line  that  can  be  drawn  from  the 
point  0  to  the  line  BC  (by  52),  any  other  line  as  OD  will  be 
longer  than  OA,  or  the  point  D  will  be  at  a  distance  from  the 
centre  greater  than  the  radius,  and  hence  is  without  the  circle. 

As  D  is  any  point  other  than  A,  A  is  the  only  point  that  is 
on  the  line  and  the  circumference,  therefore  BC  is  a  tangent 
to  the  circumference.  Q.E.D. 

151.  COR.  A  perpendicular  to  a  tangent  at  its  point  of  contact 
passes  through  the  centre  of  the  circle. 


§  152. J 


THE  CIRCLE. 


53 


EXERCISES. 

1.  Show  conversely,  a  tangent  to  the  circumference  is  perpendicular 
to  the  radius  drawn  to  the  point  of  contact. 

2.  Prove  that  the  tangents  to  a  circle  at  the  extremities  of  a  diameter 
are  parallel. 

* 

PROPOSITION  V.     THEOREM. 

152.  If  two  circumferences  intersect  each  other,  the  line  which 
joins  their  centres  is  perpendicular  to  their  common  chord  at  its 
middle  point. 


Let;  (J  and  O  be  the  centres  of  two  circumferences  which 
intersect  each  other  at  A  and  B,  and  let  the  line  CC'  intersect 
their  common  chord  at  0. 

To  prove  that  CC'  is  a  perpendicular  bisector  of  AB. 

Since  A  and  B  are  points  on  both  circles,  they  are  equally 
distant  from  C  and  also  from  C'  (by  127),  the  line  CC'  is  per- 
pendicular to  AB  at  its  middle  point  (by  56).  Q.E.D. 


EXERCISES. 

1.    If  two  circumferences  are  tangent  to  each  other,  the  straight  line 
joining  their  centres  passes  through  the  point  of  contact. 

SUGGESTION.     Draw  a  common  tangent. 


54  PLANE  GEOMETRY.  [BK.  II. 

PROPOSITION  VI.     THEOREM. 

153.  If  two  circumferences  intersect  each  other,  the  distance 
between  their  centres  is  less  than  the  sum  and  greater  than  the 
difference  of  the  radii. 

A 


Let  0  and  0'  be  two  circles  which  intersect  each  other  at  A 
and  B. 

To  prove  that  the  distance  between  their  centres  is  less  than 
the  sum  of  their  radii. 

Join  AO'  and  AO. 

Then,  in  the  triangle  O'AO, 

00'  <  A0!  +  AO  (by  77).  Q.B.D. 

Also  (by  78),  00'  >  AO  -AO'. 

154.  COR.  1.     If  the  distance  of  the  centres  of  two  circles  is 
greater  than  the  sum  of  their  radii,  they  are  wholly  exterior  to 
each  other. 

155.  COR.  2.     If  the  distance  of  the  centres  of  two  circles  is 
equal  to  the  sum  of  the  radii,  they  are  tangent  externally. 

156.  COR.  3.     If  the  distance  of  the  centres  is  less  than  the 
sum  and  greater   than   the   difference  of  the   radii,  the  circles 
intersect. 

157.  COR.  4.     If  the  distance  of  the  centres  is  equal  to  the 
difference  of  the  radii,  the  circles  are  tangent  internally. 

158.  COR.  5.     If  the  distance  of  the  centres  is  less  than  the 
difference  of  the  radii,  one  circle  is  wholly  within  the  other. 


§  160.] 


THE  CIRCLE. 


55 


PROPOSITION  VII.     THEOREM. 

159.    The  two  tangents  to  a  circumference  from  an  outside 
point  are  equal. 


Let  AB  and  AC  be  the  tangents  from  the  point  A  to  the 
circumference  whose  centre  is  at  0. 

To  prove  that  AB  =  AC. 

Draw  the  radii  OB  and  OC  and  join  AO. 

In    the    triangles    ABO  and   .400,   /.C=^B   (by   150), 
BO  =  OC,  both  being  radii,  and  the  side  AO  is  common. 

Therefore  (by  92)  they  are  equal  in  all  their  parts, 
or  AB  =  AC.  Q.E.D. 

160.   COR.     The  line  OA  bisects  the  angle  BAC,  the  angle 
BOG,  and  the  arc  BC. 

EXERCISES. 

1.  The  straight  line  drawn  from  the  centre  of  a  circle  to  the  point  of 
intersection  of  two  tangents  bisects  at  right  angles  the  chord  joining  their 
points  of  contact. 

2.  Show  that  the  sum  of  two 
opposite  sides  of   a  circumscribed 
quadrilateral  is  equal  to  the  sum 
of  the  other  two  sides. 

3.  The    bisector    of    the    angle 
between  two  tangents  to  a  circum- 
ference passes  through  the  centre. 

4.  If  tangents  are  drawn  to  a 
circumference  at  the  extremities  of 
any  pair  of  diameters,   the   figure 
thus  formed  is  a  rhombus. 

SUGGESTION.     See  151,  Ex.  2. 


56  PLANE  GEOMETRY.  [BK.  II. 

ON    MEASUREMENT. 

161.  Ratio  is  the  relation  with,  respect  to  magnitude  which 
one  quantity  bears  to  another  of  the  same  kind,  and  is  ex- 
pressed by  writing  the  first  quantity  as  the  numerator  and  the 
second  as  the  denominator  of  a  fraction. 

Thus  the  ratio  of  a  to  b  is  -;  it  is"  also  expressed  a :  b. 

The  numerical  value  of  a  ratio  is  -the  quotient  obtained  by 
dividing  the  numerator  by  the  denominator. 

162.  To  measure  a  quantity  is  to  find  its  ratio  to  another 
quantity  of  the  same  kind  called  the  unit  of  measure. 

163.  The  number  which  expresses  how  many  times  a  quan- 
tity contains  the  unit,  prefixed  to  the  name   of  the  unit,  is 
called  the  numerical  measure  of  that  quantity ;  as  5  yards,  etc. 

164.  Two  quantities  are  commensurable  when  they  have  a 
common  measure ;  that  is,  when  there  is  some  third  quantity 
of  the  same  kind  which  is  contained  an  exact  number  of  times 
in  each. 

A  B 


D 


F 


Thus  if  EF  is  contained  in  AB  3  times  and  in  CD  2  times, 
then  AB  and  CD  are  commensurable,  and  EF  is  a  common 
measure. 

165.  Two  quantities  are  incommensurable  when  they  have 
no  common  measure.  The  ratio  of  such  quantities  is  called  an 
incommensurable  ratio.  This  ratio  cannot  be  exactly  expressed 
in  figures ;  but  its  numerical  value  can  be  obtained  approxi- 
mately as  near  as  we  please. 


§  167.]  THE  CIRCLE.  57 

Thus,  suppose  G  and  H  are  two  lines  whose  ratio  is  V2. 
We  cannot  find  any  fraction  which  is  exactly  equal  to  V2  but 
by  taking  a  sufficient  number  of  decimals  we  may  find  V2  to 
any  required  degree  of  approximation. 

Thus  V2  =  1.4142135  •••, 

and  therefore  V2  >  1.414213  and  <  1.414214. 

That  is,  the  ratio  of  G  to  H  lies  between  |*-i  * fif  and  |£i£f  £|, 
and  therefore  differs  from  either  of  these  ratios  by  less  than 
one-millionth.  And  since  the  decimals  may  be  continued  with- 
out end  in  extracting  the  square  root  of  2,  it  is  evident  that 
this  ratio  can  be  expressed  as  a  fraction  with  an  error  less  than 
any  assignable  quantity. 

166.  And  in  general,  if  the  approximate  numerical  value  of 
the  ratio  of  two  incommensurable  quantities  is  desired  within 

-,  let  the  second  quantity  be  divided  into  n  equal  parts,  and 
n 

suppose  that  one  of  these  parts  is  contained  between  m  and 
m  -f  1  times  in  the  first  quantity. 

Then  the  numerical  value  of  the  ratio  of  the  first  quantity 

to  the  second  is  between  —  and  — — — ;  that  is,  the  approxi- 

n  n 

ATI 

mate  numerical  value  of  the  ratio  is  — ,  correct  within  — 

n  n 

And  since  n  can  be  taken  as  great  as  we  please,  -  is  made 

n 

correspondingly  small,  or  until  it  becomes  less  than  any  assign- 
able value,  though  it  can  never  reach  zero,  or  absolute  nothing. 

THE  METHOD   OF   LIMITS. 

167.  A  Variable  Quantity,  or  simply  a  Variable,  is  a  quan- 
tity, which  under  the  conditions  imposed  upon  it,  may  assume 
an  indefinite  number  of  values. 


58  PLANE  GEOMETRY.  [BK.  II. 

168.  A   Constant  is  a  quantity  which  remains  unchanged 
throughout  the  same  discussion. 

169.  If  a  variable  approaches  nearer  and  nearer  a  constant 
so  that  the  difference  between  it  and  the  constant  can  become 
less  than  any  quantity  that  may  be  assigned,  then  that  con- 
stant is  called  the  Limit  of  the  variable. 

170.  Suppose   a  point  M''    B 


to  move  from  A  toward  B,  under  the  conditions  that  the  first 
second  it  shall  move  one-half  the  distance  from  A  to  B  ;  that 
is,  to  M-,  the  next  second,  one-half  the  remaining  distance; 
that  is,  to  M'  ;  the  next  second,  one-half  the  remaining  dis- 
tance ;  that  is,  to  M  "  ;  and  so  on  indefinitely. 

Then  it  is  evident  that  the  moving  point  may  approach  as 
near  to  B  as  tve  please,  but  icill  never  arrive  at  B;  that  is,  the 
distance  AB  is  the  limit  of  the  space  passed  over  by  the  point. 

171.  THEOREM.  If  two  variables  are  always  equal  and  each 
approaches  a  limit,  then  the  two  limits  are  equal. 

AM  C    B 

i  _  i  _  i       i 

M<  B 


Let  AM  and  A' M'  be  two  equal  variables  which  approach 
indefinitely  the  limits  AB  and  A'B'  respectively. 

To  prove  that  AB  =  A'B'. 

If  possible,  suppose  AB  >  A'B',  and  lay  off  AC  =  A'B'. 

Then  the  variable  AM  may  assume  values  between  AC  and 
AB,  while  the  variable  AM'  is  restricted  (by  169)  to  values 
less  than  AC;  which  is  contrary  to  the  hypothesis  that  the 
variables  should  always  be  equal. 

Hence  AB  cannot  be  >  A'B',  and  in  like  manner  it  may  be 
proved  that  AB  cannot  be  <  A'B' ;  therefore  AB  =  A'B'. 

172.  COR.  If  two  variables  are  in  a  constant  ratio,  their  limits 
are  in  the  same  ratio. 


x 


I 

Let  x  and  y  be  two  variables,  so  that  -  =  m. 

y 


§  173.]  THE   CIRCLE.  59 

To  prove  that  their  limits  have  the  same  ratio. 
Now  let  x  approach  the  limit  x',  and  y  the  limit  y'. 
Then  since  the  variables  x  and  my  are  always  equal  (by  171), 
their  limits  are  equal ;  that  is,  x'  =  my'. 

x' 
Therefore  —  =  m. 

y 

MEASUREMENT   OF  ANGLES. 
PROPOSITION  VIII.     THEOREM. 

173.  In  the  same  circle,  or  in  equal  circles,  angles  at  the  centre 
are  in  the  same  ratio  as  their  intercepted  arcs. 

CASE  I.  When  the  arcs  are  com- 
mensurable. 

In  the  circle  0,  let  AOB  and  BOO 
be  two  angles  at  the  centre  intercepting 
the  commensurable  arcs  AB  and  BC. 

To  prove  that 


Let  AD  be  the  common  measure  of  the  arcs  AB  and  BC, 
and  by  applying  it  to  the  arcs  it  is  found  that  AB  contains  it 
3  times,  and  BC  4  times. 

mt,      £  sucAB     3 

1  heretore —  — 

arc  BC     4 

If  radii  be  drawn  from  the  several  points  of  division,  they 
will  divide  the  angle  AOB  into  3  parts  which  (by  145)  are 
equal,  and  BOC  into  4  parts  which  are  equal. 

mv      f  /.AOB     3 

Therefore  -  —  — 

/LBOC     4 

OON  ZAOB     axcAB 

Hence  (by  28)  ___  =  _-  Q.E.D. 


60 


PLANE   GEOMETRY. 


[BK.  II. 


CASE  II.     When  the  arcs  are  incommensurable.    If  the  arcs 
AB  and   BC  are   incommensur- 
able, cut  off  a  portion  CC'  which 
will  have  a  common  measure  with 
AB. 

Then  (by  173) 

Z.AOB 

Z  COC' 

By  taking  a  smaller  measure, 
an  arc  CC"  may  be  found  which 
is  commensurable  with  AB,  which 

would  give 

Z.AOE       arc  AB 


Z  COG"      arc  CC' 

Now  CB  is  the  limit  of  the  arc,  and  Z  COB  is  the  limit  of 
the  angle,  therefore  since  the  ratio  of  the  angles  is  equal  to  the 
ratio  of  the  arcs  at  different  stages  of  their  variation,  then 
(by  171)  their  limits  will  have  the  same  ratio,  that  is, 


Z.AOB     arc  AB 


Q.E.D. 


Z  BOC     arc  BC 

174.  SCHOLIUM.     Since  the  angle  at  the  centre  of  a  circle 
and  its  intercepted  arc  increase  and  decrease  in  the  same  ratio, 
it  is  said  that  an  angle  at  the  centre  is  measured  by  its  inter- 
cepted arc. 

PROPOSITION  IX.     THEOREM. 

175.  An  inscribed  angle  is  measured  by  one-half  the  arc  inter- 
cepted between  its  sides. 

In  the  circle  0,  let  BAG  be  an  inscribed  angle. 
To  prove  that  Z  BAG  is  measured  by  ^  arc  BC. 
Draw  the  diameter  AD  and  the  radii  OB  and  0(7. 
Since  OB  and  OA  are  radii,  the  triangle  OBA  (by  68)  is 
isosceles,  and  (by  93)  Z  B  =  Z  BAO. 


§  177.] 


THE  CIRCLE. 


61 


But  Z  BOD  being  an  exterior  angle,  it  is  equal  (by  80)  to  the 
sum  of  the  interior  and  opposite  angles,  B  and  BAO;  that  is, 

Z  BOD  =  Z  £  +  Z  BAO 

=  2  Z  BAO, 
or  Z  BAO  =  ?T  Z  BOD. 


But  Z  _BOZ)  is  measured  by  the  arc  BD  (by  174). 
Therefore  Z  .ZL4Z)  is  measured  by  |  arc  BD. 
Likewise  Z  DAG  is  measured  by  |-  arc  DC,   or  Z  iL4(7  is 
measured  by  \  arc  BDC.  Q.E.D. 


176.  COR.  1.  ^4ZZ  angles  inscribed  in 
the  same  segment  are  equal;  for  each 
is  measured  by  one-half  the  same  arc 
AFB. 


177.  COR.  2.  Every  angle  AHB, 
inscribed  in  a  semicircle,  is  a  right 
angle;  for  it  is  measured  by  one-half 
a  semi-circumference,  or  by  a  quad- 
rant. 


62 


PLANE  GEOMETRY. 


[BK.  II. 


178.  COR.  3.  Every  angle  BAG,  in- 
scribed in  a  segment  greater  than  a  semi- 
circle, is  an  acute  angle;  for  it  is  measured 
by  one-half  the  arc  BDC,  which  is  less 
than  a  quadrant. 


livery  angle  BDC,  inscribed  in  a  segment  less  than  a  semicircle, 
is  an  obtuse  angle;  for  it  is  measured  by  one-half  the  arc  BAC, 
which  is  greater  than  a  quadrant. 

An  angle  formed  by  two  chords  which 
intersect  within  a  circle,  is  measured  by 
one-half  the  sum  of  the  arcs  intercepted 
between  its  sides  and  between  its  sides  pro- 
duced. That  is,  Z  CAB  is  measured  by 
one-half  (BC  +  DE).  See  (80). 


PROPOSITION  X.     THEOREM. 

179.   An  angle  formed  by  a  tangent  and  a  chord  is  measured 
by  one-half  its  intercepted  arc. 


Let  AE  be  tangent  to  the  circumference  BCD  at  B,  and  let 
BC  be  a  chord. 

To  prove  that  Z  ABC  is  measured  by  J  arc  BC, 


§  180.]  THE   CIRCLE.  63 

At  B  erect  a  perpendicular ;  then  (by  151)  it  will  be  a 
diameter,  and  the  angle  ABD  is  a  right  angle. 

Since  a  right  angle  is  measured  by  a  quadrant  or  one-half  a 
semicircle,  Z  ABD  is  measured  by  1  arc  BCD. 

And  (by  175)  Z  CBD  is  measured  by  \  arc  CD. 

But  Z  ABC  =  Z  ABD  -  Z  CBD. 

Therefore  Z  ABC  is  measured  by  J  arc  BCD  -  -  \  arc  CD, 
or  i  (5C£>  -  CD)  =  ^arc  5(7. 

.-.  Z  ^45(7  is  measured  by  1  arc  5(7.  Q.E.D. 

PROPOSITION  XI.     THEOREM. 

180.  An  angle  formed  by  two  secants,  intersecting  witliout  the 
circumference,  is  measured  by  one-half  the  difference  of  the  inter- 
cepted arcs. 

Let  the  angle  BAG  be  formed  by  the  secants 
AB  and  AC. 


To  prove  that  the  angle  BAC  is  measured 
by  one-half  the  arc  BC  minus  one-half  the 
arc  DE. 

Join  DC. 

Then  (by  80)     Z  BDC  =  Z  C  +  Z  DAC  (or  BAC). 
By  transposing,       Z  5£><7  -  Z  C  =  Z  BAC. 
But          Z  5Z>Gy  is  measured  (by  175)  by  |  arc 
and  Z  (7  is  measured  (by  175)  by  -i-  arc 

Therefore 

Z  iL4(7  is  measured  by  -J-  arc  5(7  —  ^  arc 

EXERCISES. 

1.  An  angle  formed  by  a  tangent  and  a  secant  is  measured  by  one- 
half  the  difference  of  the  intercepted  arcs. 

2.  The  angle  formed  by  two  tangents  is  measured  by  one-half  the 
difference  of  the  intercepted  arcs. 

3.  If  a  quadrilateral  be  inscribed  in  a  circle,  the  sum  of  each  pair  of 
opposite  angles  is  two  right  angles. 


PLANE  GEOMETRY.  [BK.  II. 


PROPOSITION  XII.     THEOREM. 

181.    Two  parallel  lines  intercept  upon  the  circumference  equal 
arcs. 


Let  AC  and  BF  be  two  parallel  chords. 
To  prove  that  they  intercept  equal  arcs ;  that  is,  arc  BO  = 
arc  AF. 

Join  AB. 

Z  BAC  =  Z  ABF  (by  62). 

But          Z  BAC  is  measured  by  \  arc  BC  (175), 
and  Z  ABF  is  measured  by  1  arc  .AF. 

Since  the  angles  are  equal,  their  measures  are  equal ;  that  is, 

i  arc  BC  =  ^  arc  AF, 
or  (by  28)  arc  BC  =  arc  AF.  Q.E.D. 

EXERCISE, 

1.  Show  that  the  above  theorem  is  true  if  both  lines  are  tangents,  also 
when  one  is  a  chord  and  the  other  a  tangent. 

CONSTRUCTION. 

Up  to  the  present  time  it  has  been  assumed  that  any  needful 
line  or  combination  of  lines  could  be  drawn,  and  the  question 
has  not  arisen  as  to  the  possibility  of  drawing  these  lines  with 
accuracy. 


§  182.]  THE   CIRCLE.  65 

In  order  to  show  that  any  required  combination  of  lines, 
angles,  or  parts  of  lines  or  angles  fulfilled  the  required  con- 
ditions, principles  were  needed  long  before  they  could  be 
demonstrated. 

Sufficient  progress  has  now  been  made  to  render  it  possible 
to  show  that  every  assumed  construction  can  be  synthetically 
effected  and  proof  furnished  that  each  step  is  legitimate. 

The  only  instruments  that  can  be  employed  in  Elementary 
Geometry  are  the  ruler  and  compasses.  The  former  is  used 
for  drawing  or  producing  straight  lines,  and  the  compasses  for 
describing  circles  and  for  the  transference  of  distances. 

The  warrant  for  the  use  of  these  instruments  is  found  in 
the  three  postulates  already  given  (26). 

PROBLEMS   IN  CONSTRUCTION. 
PROPOSITION  XIII.     PROBLEM. 

182.   At  a  given  point  in  a  straight  line  to  erect  a  perpendicu 
lar  to  that  line. 


AD  C  E     B 

Let  C  be  the  given  point  in  the  line  AB. 

To  erect  a  perpendicular  to  AB  at  C. 

It  is  known  (by  54)  that  every  point  that  is  equally  distant 
from  the  extremities  of  a  straight  line  is  in  the  perpendicular 
bisector  of  that  line. 

Therefore  it  is  simply  necessary  to  make  C  the  middle  point 
of  a  portion  of  AB,  by  measuring  off  a  distance  CE,  less  than 
CB,  and  taking  CD  equal  to  CE. 


66 


PLANE   GEOMETRY. 


[BK.  II. 


To  find  another  point  equally  distant  from  D  and  E,  take 
any  radius  greater  than  DC  and  draw  arcs,  first  with  D  as  a 


-X  ,,' 

»'  % 


A    D  C  E     B 

centre,  then  with  E  as  a  centre,  and  these  arcs  will  intersect  at 
some  point,  say  F. 

Draw  FCj  and  it  will  be  the  perpendicular  required,  since  0 
and  F  are  equally  distant  from  the  points  D  and  E. 

PROPOSITION  XIV.     PROBLEM. 
183.    To  bisect  a  given  finite  straight  line. 

\'c 


\ 


Given,  the  line  AB. 

To  bisect  AB. 

It  is  known  (by  54)  that  every  point  that  is  equally  distant 
from  the  extremities  of  a  straight  line  is  in  the  perpendicular 
bisector  of  that  line. 

Therefore  it  is  necessary  to  find  two  or  more  points  equally 
distant  from  A  and  B. 

Since  radii  of  equal  circles  are  equal,  it  is  suggested  that  A 
and  B  be  made  centres  of  circles  of  equal  radii ;  then  all  points 


§  184.]  THE  CIRCLE.  67 

that  are  common  to  the  two  circles  will  be  equally  distant  from 
A  and  B,  and  hence  be  on  the  bisector  of  AB. 

With  A  as  a  centre  and  a  radius  manifestly  greater  than 
one-half  of  AB  describe  an  arc,  and  with  B  as  a  centre  describe 
an  arc  intersecting  the  former  arc  (by  156)  at  two  points,  say 
C  and  D. 

Join  CD,  and  it  will  be  the  bisector  required. 

PROPOSITION  XV.     PROBLEM. 

184.  From  a  point  without  a  straight  line,  to  let  fall  a  perpen- 
dicular upon  that  line. 

C 


B 


Let  AB  be  a  given  straight  line,  and  C  a  given  point  with- 
out the  line. 

To  let  fall  a  perpendicular  from  C  to  the  line  AB. 

If  a  line  through  C  is  to  be  perpendicular  to  AB,  it  must 
have  at  least  two  points  in  it  that  are  equally  distant  from 
two  points  in  the  line  AB. 

Let  C  be  one  of  the  former  points ;  then,  by  drawing  an  arc 
of  a  circle  with  C  as  a  centre  and  with  a  radius  manifestly 
greater  than  the  distance  from  C  to  the  line  AB,  this  arc  will 
intersect  AB  in  two  points,  say  H  and  K. 

Therefore  H  and  K  are  equally  distant  from  C. 

Another  point  equally  distant  from  H  and  K  will  be  on  the 
bisector  of  HK,  therefore  bisect  (by  183)  HK,  and  let  m  be 
the  point  of  bisection. 

Therefore  C  and  m  are  two  points  equally  distant  from  H 
and  K,  and  the  line  Cm  is  the  perpendicular  required. 


68 


PLANE   GEOMETRY. 


[BK.  II. 


EXERCISES. 

1.  From  the  extremity  of  a  straight  line  to  erect  a  perpendicular  to 
that  line. 

SUGGESTION.    Take  any  length  CD,  bisect 

it  perpendicularly;    take    C  as   a   centre,  /  \|.D' 

draw  arc  intersecting  EO  in  0  ;  with  0  as 
a  centre  and  OC  radius,  describe  circum-  Q 

ference  ;  draw  DOD' ;.  join  D'C. 

2.  Divide  a  line  into  four  equal  parts.          ,. 

TV  F  '' u 

3.  Given  the   base  and   altitude  of  an 

isosceles  triangle,  to  construct  the  triangle. 

See  94,  183. 

4.  Given  the  side  of  an  equilateral  triangle,  to  construct  the  triangle. 

PROPOSITION  XVI.  .  PROBLEM. 
185.    To  bisect  a  given  arc. 


Let  AB  be  the  given  arc. 

To  bisect  AB. 

It  is  known  (from  147)  that  the  perpendicular  bisector  of  a 
chord  is  also  a  bisector  of  the  arc  which  it  subtends. 

Therefore  draw  the  chord  AB  and  (by  183)  bisect  the  chord, 
and  the  bisector  CD  will  bisect  the  arc,  say  at  E. 

PROPOSITION  XVII.     PROBLEM. 

186.    To  bisect  a  given  angle. 

Let  A CB  be  the  given  angle. 
To  bisect  Z  ACB. 


§  187.]  THE   CIRCLE.  69 


It  is  known  (from  147)  that  the  bisector  of  an  arc  also 
bisects  the  angle  which  it  subtends. 

Therefore  draw  the  arc  DE,  and  (by  185)  bisect  DE,  say  at 
G;  then  Z  DOB  is  bisected  by  CG. 

i 

PROPOSITION  XVIII.     PROBLEM. 

187.  At  a  given  point  in  a  given  straight  line  to  construct  an 
angle  equal  to  a  given  angle. 


Let  C'  be  the  given  point  in  the  given  line  C'B',  and  C  the 
given  angle. 

To  construct  at  C'  an  angle  equal  to  Z  ACB. 

It  is  known  (from  144)  that  in  equal  circles  equal  arcs  sub- 
tend equal  angles  at  the  centre. 

Therefore  draw,  with  C  as  a  centre  and  with  C'  as  a  centre, 
equal  circles  (or  arcs) ;  then,  from  B',  measure  off  an  arc  equal 
to  arc  AB  by  taking  B'  as  a  centre,  and  with  a  radius  equal  to 
BA  draw  an  arc  intersecting  arc  B'F  at  a  point,  say  A',  and 
join  AC1 ;  then  Z  A  C'B'  =  Z  ACB. 


70  PLANE    GEOMETRY.  [BK.  II. 

EXERCISES. 

1.  To  construct  a  right  triangle,  given  an  acute  angle  and  the  base ; 
given  an  acute  angle  and  the  hypotenuse. 

2.  To  divide  an  angle  into  four  equal  parts. 

3.  Given  an  angle,  to  construct  its  complement ;  to  construct  its  sup- 
plement.    See  40,  41. 

PROPOSITION  XIX.     PROBLEM. 
188.    Given  two  angles  of  a  triangle,  to  find  the  third. 


E 

Let  A  and  B  be  the  given  angles. 

To  find  the  third  angle. 

It  is  known  (from  79)  that  the  three  angles  of  a  triangle  are 
equal  to  two  right  angles. 

It  is  also  known  (from  46)  that  the  sum  of  the  angles 
around  a  point  on  one  side  of  a  straight  line  is  equal  to  two 
right  angles. 

Therefore,  if  the  two  angles  be  added  together  so  that  their 
vertices  may  coincide  and  both  fall  on  the  same  side  of  a 
straight  line,  then  the  remaining  angle  on  that  side  will  be  the 
angle  required. 

Hence  at  a  point,  say  E  in  the  line  CD,  construct  (by  187) 
an  angle,  say  CEO-  equal  to  Z  B,  and  an  angle  FED  equal  to 
Z  Ay  then  the  remaining  Z  GEF  will  be  the  third  angle  re- 
quired. 

EXERCISES. 

1.  Given  the  base  and  vertical  angle  of  an  isosceles  triangle,  to  con- 
struct the  triangle. 

2.  Given  the  altitude  and  one  of  the  equal  angles  of  an  isosceles  tri- 
angle, to  construct  the  triangle. 


§  190'.]  THE  CIRCLE.  71 


PROPOSITION  XX.     PROBLEM. 

189.  Through  a  given  point  to  draw  a  straight  line  parallel  to 
a  given  straight  line.  A  p 

Let  BC  be  the  line  and  A  the  point.  / 

To  draw  through  A  a  line  parallel  to  BC. /L 

It  is  known  (from  62)  that  if  one  line    B         D  C 

intersect  two  other  lines  so  as  to  make  the  interior  and  op- 
posite angles  equal,  the  lines  will  be  parallel. 

Therefore,  draw  a  line  from  A  to  any  point  in  BC,  say  D, 
making  ADC  one  interior  angle. 

Then  construct  at  A  on  the  line  DA  an  angle  opposite 
Z  ADC  and  (by  187)  equal  to  it,  that  is,  the  angle  DAE ;  then 
EF  will  be  parallel  to  BC,  and  will  pass  through  A  as  re- 
quired. 

EXERCISE. 


1.    Through  a  given  point  without  a  straight  line  to  draw  a  line  mak- 
ing a  given  angle  with  that  line.  -p 

SUGGESTION.     Through  P  draw  a  line 
parallel  to  BC,  then  see  62.  B_ 

PROPOSITION  XXI.     PROBLEM. 

190.    Given  two  sides  and  the  included  angle  of  a  triangle,  to 
construct  the  triangle. 


Let  m  and  n  be  the  given  sides,  and  A'  their  included  angle. 
To  construct  the  triangle. 
Draw  a  line  AB  equal  to  m. 


72  PLANE   GEOMETRY.  [BK.  II. 

Construct  (by  187)  at  A  an  angle  BAC  equal  to  Z  A,  and 
measure  off  on  the  side  AD  a  part  equal  to  n,  and  join  CB. 

Then  ACB  will  be  the  triangle  required,  having  two  sides 
and  the  included  angle  given;  no  triangle  differing  from  it 
could  be  constructed  with  these  parts  (86). 


n 


EXERCISES. 

1.  Given  a  side  and  two  adjacent  angles  of  a  triangle,  to  construct  the 
triangle. 

2.  Given  a  side  and  any  two  angles  to  construct  the  triangle. 

3.  Show  when  the  problem  (190)  is  impossible. 

PROPOSITION  XXII.     PROBLEM. 

191.    Given   the   three   sides  of  a   triangle,   to   construct  the 
triangle. 

m 


n 


p 


Let  m,  n,  and  p  be  the  given  sides. 

To  construct  the  triangle. 

Lay  off  AB  equal  to  m ;  then  since  the  other  vertex  of  the 
triangle  must  be  at  a  distance  n  from  A,  it  will  lie  on  the  cir- 
cumference whose  centre  is  at  A  and  whose  radius  is  n ; 
therefore  draw  such  a  circle  or  a  portion  of  it. 

Likewise  the  same  vertex  must  be  at  a  distance  of  p  from  B ; 
therefore  it  will  lie  on  the  circumference  whose  centre  is  B  and 
radius  p. 


§192.]  THE  CIRCLE.  73 

Draw  such  a  circle,  and  where  the  two  circles,  or  arcs,  inter- 
sect will  be  the  vertex  C  required. 

Then  the  triangle  ABC  will  have  its  sides  equal  to  m,  n,  and 
p,  and  no  triangle  differing  from  it  could  have  the  same  three 

sides  (91). 

EXERCISES. 

1.  When  is  this  problem  impossible  ? 

2.  Two  sides  of  a  triangle  and  the  angle  opposite  one  of  them  being 
given,  to  construct  the  triangle. 

PROPOSITION  XXIII.     PROBLEM. 

192.  Given  two  adjacent  sides  and  the  included  angle  of  a 
parallelogram,  to  construct  the  parallelogram. 

With  the  sides  a,  5,  and  the  Z  A 
the  given  angle,  to   construct  the  ^ 

parallelogram. 

Lay  off  AB  equal  to  a,  construct  / 

(by  187)  the  angle  BAG  equal  to  ^L 

Z  A,  and  make  AC  =  b. 

Since    the    opposite    sides   of    a          Cfi 
parallelogram  (by  106)  are   equal,         b/  / 

the  fourth  vertex  must  be  as   far 
from  C  as  B  is  from  A ;  therefore,    _, 
it  will  be  on  a  circumference  whose 

centre  is  at  C  and  whose  radius  is  equal  to  a.  Likewise,  this 
vertex  will  be  on  a  circumference  whose  centre  is  at  B  and 
radius  equal  to  b. 

Hence  if  this  vertex  is  on  both  the  circumferences  named,  it 
will  be  at  their  intersection,  say  D. 

Join  DC  and  DB,  and  ABDC  will  be  the  parallelogram  re- 
quired. 

EXERCISES. 

1.  Construct  a  square  upon  a  given  straight  line. 

2.  Given  two  diagonals  of  a  parallelogram  and  their  included  angle  to 
construct  the  parallelogram. 


74 


PLANE  GEOMETRY. 


[BK.  II. 


PROPOSITION  XXIV.     PROBLEM. 
193.    To  inscribe  a  circle  in  a  given  triangle. 


Let  ABC  be  the  given  triangle. 

To  inscribe  a  circle  in  ABC. 

It  is  known  (from  96)  that  the  point  in  which  the  bisectors 
of  the  angles  of  a  triangle  meet  is  equally  distant  from  the 
three  sides  of  the  triangle. 

Therefore,  if  this  point  be  taken  as  a  centre,  and  the  distance 
from  it  to  any  one  side  be  used  as  a  radius,  the  circle  so 
described  will  touch  all  three  sides,  or  be  inscribed  in  the 
triangle. 

Hence  bisect  (by  186)  any  two  of  the  angles  of  the  triangle, 
and  the  point  of  intersection,  say  0,  will  be  the  centre,  and  the 
perpendicular  OM  the  radius. 

If  the  sides  of  a  triangle  are  produced  and  the  exterior 
angles  are  bisected,  the  intersections  of  the  bisectors  are  the 
centres  of  three  circles,  each  of  which  is  tangent  to  one  side  of 
the  triangle  and  the  other  two  sides  produced.  These  three 
circles  are  called  escribed  circles. 

EXERCISES. 

1.  To  draw  an  escribed  circle. 

2.  Given  the  middle  point  of  a  chord  in  a  given  circle,  to  draw  the 
chord. 

3.  Construct  an  angle  of  60°,  one  of  120°,  and  one  of  45°. 


§  105.] 


THE   CIECLE. 


75 


PROPOSITION  XXV.     PROBLEM. 

194.  To  circumscribe  a  circle  about  a  given  triangle. 

Let  ABC  be  the  triangle. 

To  circumscribe  a  circle  about  ABC. 

It  is  known  (from  54)  that  every  point 
that  is  equally  distant  from  any  two  points 
is  on  the  perpendicular  bisector  of  the  line 
joining  these  two  points. 

Therefore,  any  circle  whose  circumference 
passes  through  A  and  B  must  have  its  centre  on  the  perpen- 
dicular bisector  of  AB. 

Likewise,  the  circle  whose  circumference  passes  through  A 
and  C  must  have  its  centre  on  the  perpendicular  bisector  of  AC. 

Therefore  (by  183),  bisect  AB  and  AC]  then  the  point  in 
which  the  bisectors  meet,  say  0,  will  be  equally  distant  from 
A,  B,  and  C,  or  will  be  the  centre  of  the  circumscribing  circle. 

EXERCISES. 

1.  Through  three  points,  not  in  a  straight  line,  to  draw  a  circle. 

2.  To  circumscribe  a  circle  about  a  given  rectangle. 

PROPOSITION  XXVI.     PROBLEM. 

195.  At  a  given  point  in  a  given  circumference,  to  draw  a  tan- 
gent to  the  circumference. 

Let  0  be  the  given  circle  and  A  the  point 
on  its  circumference. 

To  draw  a  tangent  through  A. 

It  is  known  (from  150)  that  a  line  per- 
pendicular to  a  radius  at  its  extremity  is  a 
tangent  to  the  circumference. 

Therefore,  draw  the  radius  OA,  and  erect 
(by  182)  a  perpendicular  to  OA  through  A,  and  it,  say  CB,  will 
be  the  tangent  required. 


76 


PLANE  GEOMETRY. 


[BK.  II. 


EXERCISES. 

1.  On  a  given  straight  line,  to  describe  a  segment  which  shall  contain 
a  given  angle. 

SUGGESTION.     On    AB    construct    (by    187) 
/.  BAD  =  Z  C ;  draw  AH  perpendicular  to  AD 
at  D  (by  184,  Ex.  1,  see  151)  ;  bisect  AB  (by     / 
183),  and  0  will  be  the  centre  (see  147)  and    \ 
AFB  the  required  arc  (see  179).  \ 

2.  Through  a  given  point  inside  of  a  circle    A 
other  than  the  centre  to  draw  a  chord  which  is 
bisected  at  that  point. 

SUGGESTION.      Find    the    centre,    draw    the 
diameter  through  the  given  point,  and  see  147. 

PROPOSITION  XXVII.     PROBLEM. 

196.    To  draw  a  tangent  to  a  given  circle  through  a  given 
point  without  the  circumference. 


Let  0  be  the  centre  of  the  given  circle,  and  A  the  given 
point  without  the  circumference. 

To  draw  through  A  a  tangent  to  the  circumference. 

It  is  known  (from  150)  that  a  radius  and  tangent  drawn  to 
its  extremity  are  perpendicular  to  each  other. 

It  is  also  known  (from  177)  that  a  diameter  subtends  a  right 
angle. 

Therefore  the  line  joining  the  point  through  which  the  tan- 
gent is  to  pass  and  the  centre  of  the  given  circle  must  sub- 
tend a  right  angle  or  be  the  diameter  of  an  auxiliary  circle. 


§  196.] 


THE  CIRCLE. 


77 


Hence  draw  AO,  bisect  it  (by  183)  at  0',  say,  then  describe  a 
circle  with  O'O  as  a  radius  and  0'  a  centre,  and  connect  A 
with  the  points  where  this  circle  intersects  the  given  circle,  say 
B  and  C,  then  AB  and  AC  will  be  the  tangents  required, 
Z  ABO  and  Z  ACO  being  right  angles. 


EXERCISES. 

1.    To  describe  a  circle  tangent  to  a  given  straight  line,  having  its 
centre  at  a  given  point. 

SUGGESTION.      See    184 
and  150. 


C 


2.  Through  a  given  point 
to  describe  a  circle  of  given  _ 
radius,  tangent  to  a  given 

straight  line.  A  D 

SUGGESTION.  Erect  DB  (=C)  _L  to  AB  at  B,  draw  DE  parallel  to 
AB,  with  P  as  centre,  and  radius  equal  to  (7,  cut  ED  in  0,  then  0  is  the 
centre. 

3.  Show  when  the  above  problem 
is  impossible. 

4.  To  describe  a  circle  touching 
two  given  straight  lines,  one  of  them 
at  a  given  point. 

SUGGESTION.     See  160,  150. 

5.  To  find  the  centre  of  a  given  arc  or  circle. 

6.  To  draw  a  tangent  common  to  two  circles,  C  and  c. 

SUGGESTION.  Draw  two  parallel  radii  CB  and  cb  ;  draw  Bb  and  con- 
tinue it  until  it  meets  Cc  produced  in  A.  See  196. 


BOOK  III. 

RATIO  AND  PROPORTION.     SIMILAR  FIGURES. 


DEFINITIONS. 

197.  A  Proportion  is  an  equality  of  ratios.      (See  161-163.) 
That  is,  if  the  ratio  of  a  to  b  is  equal  to  the  ratio  of  c  to  d, 

they  form  a  proportion,  which  may  be  written 

a  :  b  =  c  :  d.  or  -  =  -,  or  a:b:  :  c:  d, 

b      d' 

and  is  read  a  is  to  b  as  c  is  to  d. 

198.  The  four  terms  of  the  two  equal  ratios  are  called  the 
Terms  of  the  proportion.    The  first  and  fourth  terms  are  called 
the  Extremes,  and  the  second  and  third  the  Means.     Thus,  in 
the  above  proportion,  a  and  d  are  the  extremes,  and  b  and 
c  the  means. 

The  first  and  third  terms  are  called  the  Antecedents,  and  the 
second  and  fourth  the  Consequents.  Thus,  a  and  c  are  the 
antecedents,  and  b  and  d  the  consequents. 

The  fourth  term  is  called  a  Fourth  Proportional  to  the  other 
three.  Thus,  in  the  above  proportion,  d  is  a  fourth  propor- 
tional to  a,  b,  a,nd  c. 

In  the  proportion  a :  b  =  b  :  c,  c  is  a  third  proportional  to  a 
and  b,  and  &  is  a  mean  proportional  between  a  and  c. 

78 


BK.  III.,  §202.]      RATIO  AND  PROPORTION.  79 

PROPOSITION  I. 

199.  If  four  quantities  are  in  proportion,  the  product  of  the 
extremes  is  equal  to  the  product  of  the  means. 

Let  a :  b  =  c  :  d. 

To  prove  ad  =  be. 

By  definition  (197),  -  =  -• 

b     d 

Clearing  of  fractions,       ad  =  be.  Q.E.D. 

200.  COR.     If  a:b  =  b:  c, 
then  (by  199)  b2  =  ac. 

.-.  b  =  Vac.  Q.E.D. 

That  is,  the  mean  proportional  between  two  quantities  is  equal 
to  the  square  root  of  their  product. 

PROPOSITION  II.     THEOREM. 

201.  CONVERSELY,  if  the  product  of  two  quantities  is  equal  to 
the  product  of  two  others,  one  pair  may  be  made  the  extremes, 
and  the  other  pair  the  means,  of  a  proportion. 

Let  ad  =  be. 

Dividing  both  members  of  the  equation  by  bd} 

ad     be         a     c 

QT»       . 

bd     bd'        b     d 
That  is,  a  :  b  =  c  :  d.  Q.E.D. 

PROPOSITION  III.     THEOREM. 

202.  In  any  proportion  the  terms  are  in  proportion  by  Alter- 
nation ;  that  is,  the  Jirst  term  is  to  the  third  as  the  second  term 
is  to  the  fourth, 

Let  a  :  b  =  c  :  d. 

Then  (by  199)  ad  =  be. 

Whence  (by  201)  a:c  —  b:d.  Q.E.D, 


80  PLANE  GEOMETRY.  [BK.  III. 


PROPOSITION  IV. 

203.  If  four  quantities  are  in  proportion,  they  are  in  propor- 
tion by  Inversion ;  that  is,  the  second  term  is  to  the  first  as  the 
fourth  term  is  to  the  third. 

Let  a:b  =  c:d. 

To  prove  b  :  a  =  d  :  c. 

If  a:b  =  c:  d, 
then  (by  199)  bc  =  ad. 

TV*   •  j    i  be      ad 

Divide  by  ac,  —  —  — ; 

ac     ac 

,,  b      d 

that  is,  -  =  -, 

a      c 

or  6  :  a  —  d  :  c.  Q.E.D. 

PROPOSITION  V.     THEOREM. 

204.  In  any  proportion  the  terms  are  in  proportion  by  Com- 
position;  that  is,  the  sum  of  the  first  two  terms  is  to  the  first  term 
as  the  sum  of  the  last  two  terms  is  to  the  third  term. 

Let  a:b  =  c:d. 

To  prove  a  -{-b  :  a:  :  c  +  d:  c. 

If  a  :  b  =  c  :  d, 

then  (by  199)  ad  =  bc. 

Adding  both  members  of  the  equation  to  ac, 

ac  -f-  od  =  ac  +  be, 
or  a(c  +  d)  —  c(a  +  6). 

Therefore  (by  201),  a  +  6:a::c  +  d:c. 

Similarly.  a +  b  :b  : :  c  +  d  :  d.  Q.E.D. 


§206.]  RATIO  AND  PROPORTION.  81 

PROPOSITION  VI. 

205.  If  four  quantities  are  in  proportion,  they  are  in  propor- 
tion by  Division  ;  that  is,  the  difference  of  the  first  and  second  is 
to  the  first  as  the  difference  of  the  third  and  fourth  is  to  the  third. 

Let  a  :  b  =  c  :  d. 

To  prove  a  —  b  :  a  =  c  —  d:  c. 
If  a  :  b  =  c  :  d, 

then  (by  199)  ad  =  be. 

• 

Subtract  both  members  of  this  equation  from  ac,  then 

ac  —  ad  =  ac  —  be, 

or  a  (c  -  -  d)  =  c  (a  -  -  b). 

Therefore  (by  201),  a--b:a  =  c  —  d:  c. 
Similarly,  a  —  b:b  =  c  —  d:d.  Q.E.D. 

PROPOSITION  VII. 

206.  If  four  quantities  are  in  proportion,  they  are  in  propor- 
tion by  Composition  and  Division  ;  that  is,  the  sum  of  the  first 
and  second  is  to  their  difference  as  the  sum  of  the  third  and  fourth 
is  to  their  difference. 

Let  a:b  =  c:d. 

To  prove  a  +  b  \  a  —  b  =  c  +  d:  c  —  d. 


(By  204),  =  - 

b  d 


and  (by  205) 

} 

i-,-.. 
by  division, 


b  d 

a  -\-b      c  -\-  d 

-  =  -      — 

a  —  b     c  —  d 
.\  a  +  b  :  a  —  b  =  c  +  d  :  c  —  d.  Q.E.D. 


82  PLANE  GEOMETRY.  [Ex.  III. 


PROPOSITION  VIII. 

207.  The  products  of  the  corresponding  terms  of  two  or 
more  proportions  are  proportional. 

Let  a  :  b  =  c  :  d  ,  and  e:f=g:h. 

To  prove  ae  :  bf=  eg  :  dh. 

Writing  the  proportions  in  another  form, 

«  =  *,  and  1  =  2. 

b     d'  f     h 

Multiplying  these  equations  member  by  member, 

ae      eg 
bf==dti 

s 

or  ae  :  bf=  eg  :  dh.  Q.E.D. 

208.  COR.     If  the  corresponding  terms  of  the  proportions 
are  equal  ;  that  is,  if  e  =  a,  f=  &,  g  =  c,  and  h  =  d,  the  result 
of  the  preceding  theorem  becomes 

a2  :  b2  =  c2  :  d\ 

And  in  general  in  any  proportion  like  powers  of  the  terms  are 
in  proportion. 

PROPOSITION  IX.     THEOREM. 

209.  In  a  series  of  equal  ratios,  any  antecedent  is  to  its  con- 
sequent as  the  sum  of  all  the  antecedents  is  to  the  sum  of  all  the 
consequents. 

Let  a:b  =  c:  d  =  e:f. 

To  prove     a  +  c  +  e:b  +  d  +/=  a:b  =  c:  d  =  e:f. 

Let  r  be  the  value  of  the  equal  ratios,  that  is, 

a  c  -,    e 

-  =  r,   -  =  r,   and^r. 


§211.]  RATIO  AND  PROPORTION.  83 

From  these  equations, 

a=  br,    c  =  dr,    e  =fr, 

or  by  addition,  a  +  c  +  e  =  br  +  dr  +  fr 


•,..-,.  a  +  c  -f  e 

By  dividing,  -  =  r. 

b  +  d+f 

CL         C         & 

But  by  hypothesis,  r  =  -  =  -  =  -• 

m,       £  a+c+eace 

Therefore  -  =  -  =  -  =  -, 

b+d+f     b      d     f 
that  is,  a  +  c  +  e:b  +  d  +/=  a:b  =  c:d  =  e:f.      Q.E.D. 

PROPOSITION  X.     THEOREM. 

210.  In  any  proportion,  if  the  antecedents  are  multiplied  by 
any  quantity,  as  also  the  consequents,  the  resulting  terms  ivill  be 
in  proportion. 

Let  a  :  b  =  c  :  d. 

Then  -  =  -• 

b     d 

m 
Multiplying  both  members  of  the  equation  by    -, 

iL 

ma      me 


nb      nd 
That  is,  ma  :  nb  —  me  :  nd. 

T    v,  abed 

In  like  manner,  _ :  _  =  __  :  _. 

m   n      m   n  y.-E-.u. 

211.    SCHOLIUM.     Either  m  or  n  may  be  unity. 

EXERCISES. 

1.  Show  that  equimultiples  of  two  quantities  are  in  the  same  ratio  as 
the  quantities  themselves. 

2.  Show  that  if  four  quantities  are  in  proportion,  their  like  roots  are 
in  proportion. 


84  PLANE   GEOMETRY.  [BK.  IIL 

PROPORTIONAL  LINES. 

Two  straight  lines  are  said  to  be  divided  proportionally  when 
their  corresponding  segments,  or  parts, 

are  in  the  same  ratio  as  the  lines  them-      ^ 5 ? 

selves. 

Thus  the  lines  AB  and  CD  are  divided          °       f  D 

proportionally  at  E  and  F  if 

AB  :  AE  =  CD  :  CF. 

212.  When  a  finite  straight  line,  as  AB,  is  cut  at  a  point  X 
between  A  and  5,  it  is  said  to  be  divided  internally  at  X,  and 
the  two  parts  AX  and  J5X  are  called 

segments.    But  if  the  straight  line  AB      | ^     ,         ^ 

is  produced,  and  cut   at   a  point   Y    A  fi 

beyond  AB,  it  is  said  to  be  divided 

externally  at  Y}  and  the  parts  A  Y  and  B  Y  are  called  segments. 
The  given  line  is  the  sum  of  two  internal  segments,  or  the 
difference  of  two  external  segments. 

When  a  straight  line  is  divided  internally  and  externally 
into  segments  having  the  same  ratio,  it  is  said  to  be  divided 
harmonically. 

PROPOSITION  XI.     THEOREM. 

213.  A  straight  line  parallel  to  one  side  of  a  triangle  divides 
the  other  tivo  sides  proportionally. 

In  the  triangle  ABC  let  DE  be  parallel 
to  BC. 

To  prove     AD  :  DB  =  AE  :  EC. 

CASE  I.     When  AD  and  DB  are   com- 


_  p 

mensurable. 

Take  AF.  anv  common  measure  of  AD 

*/ 

and  DB,  and  suppose  it  to  be  contained  4 
times  in  AD  and  3  times  in  DB.  B 


§214.]  RATIO  AND  PROPORTION.  85 


Then 

DB      3 

Through  the  several  points  of  division  of  AB  draw  lines 
parallel  to  BC-,  then  since  these  parallels  cut  off  equal  lengths 
on  AB,  they  will  (by  110)  cut  off  equal  lengths  on  AC. 

Therefore,  AE  will  be  divided  into  4  equal  parts  and  EC 
into  3  ;  that  is, 


Hence  (by  28),  —  =  —  , 

DB     EC 

or  AD:DB  =  AE:  EC. 

CASE  II.  When  AD  and  DB  are  incom- 
mensurable. 

In  this  case  we  know  (170)  that  we  may 
always  find  a  line  AG  as  nearly  equal  as  we 
please  to  AD,  and  such  that  AG  and  DB  are 
commensurable. 

Draw  GH  parallel  to  BC;  then 


As  these  two  ratios  are  always  equal  while  the  common 
measure  is  indefinitely  diminished,  they  will  be  equal  as  GH 
approaches  DE. 

Therefore,  this  quality  of  ratios  will  exist  (by  172)  as  the 
limiting  position  DE  is  approached  ;  that  is, 


214.   COR.     By  composition  (204), 

AD  +  DB  :  AD  =  AE  +  EC  :  AE, 
or  AB:AD  =  AC:AE. 

Likewise  (by  204),     AB  :  DB  =  AC  :  EC, 
and  (by  202),  AB:AC  =  DB:  EC. 


86  PLANE  GEOMETRY.  [BK.  III. 


EXERCISES.  Q 

A 

1.  Conversely,  if  a  straight  line  divides  /  \ 

two  sides  of  a  triangle  proportionally,  it  is  ^ 

parallel  to  the  third  side. 

2.  If  two  straight  lines  AB,  CD  are  cut 
by  any  number  of  parallels,  AC,  EF,  GH, 


BD,  the  corresponding  intercepts  are  pro-         Q  I  \  H 

portional. 

SUGGESTION.     See  214.  /  \ 

B  L  _  \:D 

PROPOSITION  XII.     THEOREM. 

215.    The  bisector  of  any  angle  of  a  triangle  divides  the  oppo- 
site side  into  segments  proportional  to  the  adjacent  sides. 


Let  ABC  be  the  triangle,  and  GB  the  bisector  of  the  angle 
ABC. 

To  prove  AG  :  GC  =  AB  :  BC. 

Draw  CD  parallel  to  GB,  and  produce  AB  to  D. 

Then  (by  63)  Z  BDC  =  Z  ABG, 

and  (by  62)  Z  BCD  =  Z  GBC. 

But  by  construction,  Z  ABG  =  Z  GBC; 
therefore  (by  28)  Z  BDC  =  Z  BCD ; 

hence  (by  93)  the  triangle  BCD  is  isosceles,  and 

BC  =  BD. 
It  is  known  (from  213)  that 

AG:GC=AB:BD. 
Substituting  for  BD  its  equal  BC, 

AG:GC  =  AB:  BC.  Q.E.D. 


§  217.]  RATIO  AND  PROPORTION.  87 

EXERCISES. 

1.  If  a  line  divides  one  side  of  a  triangle  into  segments  that  are  pro- 
portional to  the  adjacent  sides,  it  bisects  the  opposite  angle. 

2.  The  bisector  of  an  exterior  angle  of  a  triangle  divides  the  opposite 
side    externally  into    segments 

proportional    to    the     adjacent 
sides. 


3.  If      (in     215), 

BC  =  6,  and  CA  =  9,  find  AG- 
and  GC. 

4.  If  AB  =  5,  BC  =  7,  and 
CA  =  8,  find  AD  and  ED. 

5.  Bisectors  of  an  interior  and  exterior  angle  at  the  vertex  of  a  triangle 
divide  the  opposite  side  harmonically. 

SUGGESTION.     See  212. 

SIMILAR  POLYGONS. 

216.  DEFINITIONS.  Two  polygons  are  called  Similar  when 
they  are  mutually  equiangular  (122)  and  have  their  homolo- 
gous sides  proportional  (122). 


E  D  Er 


That  is,  the  polygons  ABODE  and  A'B'C'D'E'  are  similar  if : 
Z  A  =  Z  A',  ZB  =  ZB',  Z  C  =  Z  C",  etc., 

AB       BC       CD 

and  • = = ,  etc. 

A'B'     B'C'      C'D1' 

217.   In  two  similar  polygons,  the  ratio  of  any  two  homolo- 
gous sides  is  called  the  Ratio  of  Similitude  of  the  polygons. 


PLANE  GEOMETRY.  [BK.  III. 


PROPOSITION  XIII.     THEOREM. 
218.    Triangles  which  are  mutually  equiangular  are  similar. 

Let  ABC  and  A'B'C'  be  two  mutu- 
ally equiangular  triangles. 

To  prove  that  ABC  and  A'B'C1  are 
similar  triangles. 

Lay  off  on  AB  a  distance  equal  to 
A'B'  and  on  AC  make  AE  equal  to 

A'C.  *  c  B ' 

The  triangles  ADE  and  A'B'C'  are  (by  86)  equal,  having  the 
included  Z  A  equal  to  Z  A,  and  the  sides  AD  and  AE  equal 
to  AB'  and  AC",  by  construction.  Therefore  Z  ^.ZXE  =  Z  5', 
but  by  hypothesis,  Z  5  =  Z  B',  hence  Z  ADJ£  =  Z  5,  there- 
fore (by  63)  DE  is  parallel  to  BC. 

If  ZXE  is  parallel  to  BC,  we  have  (by  214) 


or  substituting  for  AD  its  equal  A'B',  and  for  AE,  A'C', 

AB:A'B'  =  AC:A'C'. 

Similarly,  it  can  be  shown,  by  laying  off  on  BA  a  distance 
equal  to  B'A,  and  on  BC  a  distance  equal  to  B'C',  that 

BA:B'A'  =  BC:B'C'.  Q.E.D. 

219.  COR.  1.   Two  triangles  are  similar  when  two  angles  of 
the  one  are  equal  respectively  to  two  angles  of  the  other.   (See  82.) 

220.  COR.  2.    A  triangle  is  similar  to  any  triangle  cut  off  by  a 
line  2iarallel  to  one  of  its  sides. 

221.  SCHOLIUM.     In  similar  triangle 3  the  homologous  sides 
lie  opposite  the  equal  angles. 


§222.]  RATIO  AND  PROPORTION.  89 


PROPOSITION  XIV.     THEOREM. 

222.    Two  triangles  are  similar  when  their  homologous  sides 
are  proportional. 

A 


B  C 

In  the  triangles  ABC  and  A'B'C',  let 

AB  =  AC  =  EG 
A'B'      AC1     B'C'' 

To  prove  that  the  triangles  are  similar. 

Take  AD  =  A'B'  and  AE  =  A'C1,  and  join  DE. 

Then  from  the  given  proportion  we  have 


AD  ~  AE' 

therefore  (by  converse  of  214,  Ex.  1)  the  line  DE  is  parallel  to 
BC,  and  the  angles  ADE  and  B  having  their  sides  parallel  and 
similarly  directed  are  (by  63)  equal  ;  likewise,  Z  AED  =  Z  C. 

\ 

Hence  the  triangles  ADE  and  ABC  are  mutually  equiangu- 
lar and  (by  218)  are  similar;  that  is, 

AB      BC 


or 
r 


AD      DE'         A'B'      DE 


But,  by  hypothesis, 

A'B'     B'C' 

These  last  two  proportions  agree  term  for  term  except  the 
last  in  each,  and  these  must  therefore  be  equal,  or  B'C'  =  DE. 

Hence  the  triangles  ADE  and  A'B'C'  are  mutually  equi- 
lateral and  therefore  equal. 

But  the  triangle  ADE  has  been  proved  similar  to  ABC. 

Hence  the  triangle  A'B'C'  is  similar  to  ABC.  Q.E.D. 


90  PLANE  GEOMETRY.  [BK.  IIL 

223.  SCHOLIUM.  Two  polygons  are  similar  when  they  are 
mutually  equiangular  and  have  their  homologous  sides  propor- 
tional. But  in  the  case  of  triangles  we  learn,  from  Proposi- 
tions XIII.  and  XIV.,  that  either  of  these  conditions  involves 
the  other. 

This,  however,  is  not  necessarily  the  case  with  polygons 
of  more  than  three  sides ;  for  even  with  quadrilaterals,  the 
angles  can  be  changed  without  altering  the  sides,  or  the 
proportionality  of  the  sides  can  be  changed  without  altering 
the  angles. 

EXERCISES. 

1.  Two  right  triangles  are  similar  when  they  have  an  acute  angle  of 
one  equal  to  an  acute  angle  of  the  other. 

2.  Two  triangles  are  similar  when  they  have  an  angle  of  one  equal  to 
an  angle  of  the  other,  and  the  sides  including  these  angles  proportional. 

3.  Two  triangles  are  similar  when  the  sides  of  one  are  parallel  respec- 
tively to  the  sides  of  the  other. 

4.  Two  triangles  are  similar  when  the  sides  of  one  are  perpendicular 
respectively  to  the  sides  of  the  other. 

SUGGESTION.     See  64. 

5.  The    homologous    altitudes  of   two 
similar  triangles  have  the  same  ratio  as 
any  two  homologous  sides. 

6.  If  in  any  triangle  a  parallel  be  drawn 
to  the  base,  all  lines  from  the  vertex  will 

divide  the  base  and  its  parallel  proper-  -  — /^         /B  \C  — \Z 

tionally.  '  » 

SUGGESTION.     See  218. 

7.  Two  parallelograms  are  similar  when  they  have  an  angle  equal  and 
the  including  sides  proportional. 

8.  Two  rectangles  are  similar  when  they  have  two  adjacent  sides  pro- 
portional. 

9.  If  two  triangles  stand  upon  the  same  base,  and  not  between  the 
same  parallels,  the  figure  formed  by  joining  the  middle  points  of  their 
sides  is  a  parallelogram. 

10.  If  from  any  two  diametrically  opposite  points  on  the  circumference 
of  a  circle  perpendiculars  be  drawn  to  a  straight  line  outside  the  circle, 
the  sum  of  these  perpendiculars  is  constant. 


§224.J  RATIO  AND   PROPORTION.  91 


PROPOSITION  XV.     THEOREM. 

224.  Two  polygons  are  similar  when  they  are  composed  of  the 
same  number  of  triangles,  similar  each  to  each  and  similarly 
placed. 


D 


Let  ABODE  and  A'B'C'D'E'  be  two  polygons  composed  of 
the  same  number  of  similar  triangles  similarly  placed. 

To  prove  that  the  polygons  are  similar ;  that  is,  that  they 
are  mutually  equiangular,  and  that  their  homologous  sides  are 
proportional. 

Since  the  triangles  AEB  and  A'E'B'  are  similar  by  hypoth- 
esis, they  are  (by  223)  mutually  equiangular ;  that  gives 

Z.A  =  Z.  A',  and  Z  ABE  =  Z  A'B'E'. 

Likewise,  in  the  triangles  EBC  and  E'B'C', 


or  by  addition,  Z  ABE  +  Z  EBC  =  Z  A'B'E'  +  Z  E'B'C', 

or  Z  ASC  =  Z  A' B'C'. 

In  like  manner, 

Z  BCD=Z.  B'C'D',  Z  CDE=Z  C'D'E',  and  Z  DEA  =  ZD'E'A'. 

Since  the  triangles  are  similar,  their  homologous  sides  are 
proportional,  which  gives 

J|  =  ||,   and  Jf  =  |§. 

or  (by  28),  -^=BC. 

J  A'B'     B'C' 


92 


PLANE  GEOMETRY. 


[BK.  III. 


In  like  manner, 
AB 


BC       CD       DE       EA 


A'B'     B'C1      C'D'     D'E'     E'A 


Q.E.D 


225.  COR.     Conversely,  two  similar  polygons  may  be  divided 
into  the  same  number  of  triangles,  similar  each  to  each,  and 
similarly  placed. 

PROPOSITION  XVI.     THEOREM. 

226.  The  perimeters  of  two  similar  polygons  have  the  same 
ratio  as  any  two  homologous  sides. 


Let  the  tAvo  similar  polygons  be  ABCDE  and  A'B'C'D'E', 
and  let  P  and  P'  represent  their  perimeters. 

To  prove  P  :  P'  :  :  AB  :  A'B'. 

Since  the  polygons  are  similar  (by  223), 

AE       ED       DC       CB       BA 


A'E1  ~  ~  E'D'  "  D'  C'      C'B'  ~  '  B'A1 

The  sum  of  the  antecedents  will  have  the  same  ratio  to  the 
sum  of  the  consequents  that  any  antecedent  has  to  its  conse- 
quent (by  209)  ;  that  is, 

AE  +  ED  +  DC  +  CB  -f  BA          AE 


I  Z7V 


or, 


A'E'  +  E'D'  +  DC'  +  C'B'  +  B'A'     A'E 

P       AE 


I  17>t 


A'E 


Q.E.D. 


§227.J  PATIO   AND   PROPORTION.  93 


PROPOSITION  XVII.     THEOREM. 

227.    If  in  a  right  triangle  a  perpendicular  be  drawn  from  the 
vertex  of  the  right  angle  to  the  hypotenuse: 

I.  It  divides  the  triangle  into  two  right  triangles  which  are 
similar  to  the  ivhole  triangle,  and  also  to  each  other. 

II.  The  perpendicular  is  a  mean  proportional  between  the  seg- 
ments of  the  hypotenuse. 

III.  Each  side  of  the  right  triangle  is  a  mean  proportional 
between  the  hypotenuse  and  its  adjacent  segment. 

B 


F  * 

In  the  right  triangle  ABC,  let  BF  be  drawn  from  the  vertex 
of  the  right  angle  B,  perpendicular  to  the  hypotenuse  AC. 

1.  To  prove  that  ABF  is  similar  to  BFC,  and  that  each  is 
similar  to  ABC. 

The  triangles  ABF  and  ABC  have  the  angle  A  common, 
and  Z  AFB  =  Z  ABC,  therefore  their  third  angles  (by  82)  are 
equal. 

Hence  the  triangles  are  mutually  equiangular  and  (by  218) 
are  similar. 

Likewise,  the  triangles  BFC  and  ABC  are  similar. 

Therefore,  if  the  triangles  ABF  and  BFC  are  similar  to 
ABC,  they  will  be  similar  to  one  another. 

2.  To  prove  that  AF :  BF=  BF:  FC. 

Since  the  triangles  ABF  and  BFC  are  similar,  their  homolo- 
gous sides  are  (by  223)  proportional,  that  gives 

AF:BF  =  BF:  FC,  or  (by  199)  BF2  =  AF  x  FC. 


94 


PLANE  GEOMETRY. 


[BK.  III. 


3.    To  prove  that 

AC:BC  =  BC:  FC,  or  (by  199)  BC~  =  AC  x  FC. 

Since  the  triangles  ABC  and  FBC  are  similar,  their  homol- 
ogous sides  (by  223)  are  proportional,  which  gives 

AC:BC  =  BC:FC. 

In  a  similar  manner,  it  can  be  shown  that 
AC:AB  =  AB:AF,  or  (by  199)  AB2  =  AC  X  AF.  Q.B.D. 


228.  COB.  Since  an  angle  inscribed  in 
a  semicircle  is  a  right  angle  (177),  it  fol- 
lows that 

I.  The  perpendicular  from  any  point  in 
the  circumference  of  a  circle  to  a  diameter 

is  a  mean  proportional  betiveen  the  segments  of  the  diameter. 

II.  TJie  chord  drawn  from  the  point  to  either  extremity  of  the 
diameter  is  a  mean  proportional  between  the  whole  diameter  and 
the  adjacent  segment. 

EXERCISES. 

1.  The  squares  on  the  two  sides  of  the  right  triangle  have  the  same 
ratio  as  the  adjacent  segments  of  the  hypotenuse. 

2.  The  square  on  the  hypotenuse  has  the  same  ratio  to  the  square  on 
either  side  as  the  hypotenuse  has  to  the  segment  adjacent  to  that  side. 

3.  Two  isosceles  triangles  are  similar  when  their  vertical  angles  are 
equal. 


§232.]  RATIO  AND  PROPORTION.  95 


PROPOSITION  XVIII.     THEOREM. 

229.  If  any  ttuo  chords  are  drawn  through  a  fixed  point  in  a 
circle,  the  product  of  the  segments  of  one  is  equal  to  the  product 
of  the  segments  of  the  other. 


Let  AB  and  A'B'  be  any  two  chords  of  the  circle  ABB' 
passing  through  the  point  P. 

To  prove  that       AP  x  BP  =  A'P  x  B'P. 

Join  AA  and  BB'. 

In  the  two  triangles  APA  and  BPB',  the  vertical  angles 
A  PA  and  BPB'  are  (by  49)  equal,  Z  B'  and  Z  A  are  equal, 
both  being  measured  by  one-half  of  the  same  arc  A'B,  and 
Z  B  =  Z  A'  for  the  same  reason. 

The  triangles  are  therefore  equiangular  and  (by  218)  are 
similar,  which  gives 

A'P :  PB  =  AP :  PB',  or  (by  199)  A'P  X  PB1  =  AP  x  PB. 

230.  When  four  quantities,  such  as  the  sides  about  two 
angles,  are  so  related  that  a  side  of  the  first  is  to  a  side  of  the 
second  as  the  remaining  side  of  the  second  is  to  the  remaining 
side  of  the  first,  the  sides  are  said  to  be  reciprocally  propor- 
tional.    Therefore 

231.  COR.  1.    If  two  chords  cut  each  other  in  a  circle,  their 
segments  are  reciprocally  proportional. 

232.  COR.  2.  If  through  a  fixed  point  within  a  circle  any  num- 
ber of  chords  are  drawn,  the  products  of  their  segments  are  all  equal. 


96 


PLANE  GEOMETRY. 


[BK.  III. 


PROPOSITION  XIX.     THEOREM. 

233.  If  from  a  point  without  a  circle  a  tangent  and  a  secant 
be  drawn,  the  tangent  is  a  mean  proportional  betiveen  the  whole 
secant  and  the  external  segment. 

Let  PC  and  PB  be  a  tangent  and  a  secant  drawn  from  the 
point  P  to  the  circle  CAB. 

To  prove  that     PB:PC=PC:  PA. 

Join  CA  and  CB. 

In  the  two  triangles  PC  A  and  PCB  the 
angle  P  is  common,  and  Z  PC  A  =  Z  PBC, 
being  measured  by  one-half  of  the  same  arc 
CA ;  then  (by  82),  Z  PAG  =  Z  PCB. 

Therefore,  the  triangles  are  equiangular 
and  are  (by  218)  similar,  which  gives 

PB:PC=PC:PA;   or  (by  199),  PC*  =  PB  x  PA.    Q.E.D. 


234.   COR.     PC'  =  BP  x  PA ;  therefore  (by  28), 

or  PC=PC'. 


EXERCISES. 


1.  If  from  a  point  without  a  circle  two 
secants  be  drawn,  the  product  of  one  secant 
and  its  external  segment  is  equal  to  the  prod- 
uct of  the  other  and  its  external  segment. 

2.  If  from  a  point  without  a  circle  any 
number  of  secants  are  drawn,  the  products  of 
the  whole  secants  and  their  external  segments 
are  all  equal. 

SUGGESTION.     Draw   a   tangent    PC    and 
apply  233. 


§237.]  RATIO  AND  PROPORTION.  97 

PROPOSITION  XX.     PROBLEM. 

235.    To  divide  a  given  straight  line  into  parts  proportional  to 
any  number  of  given  lines. 


m 


Let  AB,  m,  n,  and  o  be  given  straight  lines. 

It  is  required  to  divide  AB  into  parts  proportional  to  the 
given  lines  m,  n,  and  o. 

It  is  known  (from  213)  that  lines  drawn  parallel  to  the  base 
of  a  triangle  divide  the  other  two  sides  proportionally. 

Therefore,  form  with  AB  a  triangle  by  drawing  an  indefinite 
straight  line  from  A ;  measure  off  on  this  line  a  part  equal  to 
m,  say  AC;  then  n,  say  CE,  and  o,  say  EF,  and  join  BF,  thus 
forming  the  triangle  AFB. 

Through  the  points  E  and  C  draw  lines  parallel  to  FB, 
meeting  AB  in  K  and  H;  then 

AH:  HK:KB  =  AC:  CE:EF=  mm:  o. 

236.  COR.  1.     By  making  AC=  CE  =  EF,  the  line  AB  will 
be  divided  equally. 

237.  COR.  2.     By  making  AC  =  m,  AH=n,  and  CE  =  o, 
we  would  have  (by  213),  m  :  n  :  :  o  :  UK]  that  is,  HK  would 
be  a  fourth  proportional  to  m,  ?i,  and  o. 

EXERCISE, 

To  divide  a  given  straight  line  into  three  segments,  -4,  B,  and  (7,  such 
that  A  and  B  shall  be  in  the  ratio  of  two  given  straight  lines  m  and  n, 
and  B  and  C  shall  be  in  the  ratio  of  two  other  straight  lines  o  and  p. 


98 


PLANE  GEOMETRY. 


[BK.  III. 


m- 


n- 


PROPOSITION  XXI.     PROBLEM. 

238.  To  find  a  mean  proportional  between  two  given  straight 
lines. 

Let  m  and  n  be  the  two  lines. 

To  find  a  mean  proportional  to  them. 

It  is  known  (from  228)  that  the  perpen- 
dicular from  the  circumference  to  the 
diameter  is  a  mean  proportional  between 
the  segments  of  the  diameter. 

Therefore,  if  a  diameter  be  made  of  m 
and  n  and  a  perpendicular  erected  at  their 
point  of  union,  the  portion  included  between  the  diameter  and 
the  circumference  will  be  the  mean  proportional  required; 
that  is,  lay  off  AD  =  m  and  DB  =  n.  Describe  upon  AB  as  a 
diameter  a  circle,  erect  (by  182)  a  perpendicular  at  D,  and  CD 
will  be  a  mean  proportional,  or 

AD:CD  =  CD:  DB,    or   m:CD  =  CD:  n. 

239.  The   mean   proportional   between  two   lines   is  often 
called  the  geometric  mean,  while  half  their  sum  is  called  the 
arithmetic  mean. 

PROPOSITION  XXII.     PROBLEM. 

240.  On  a  given  straight  line,  to  construct  a  polygon  similar  to 
a  given  polygon. 

Let    ABCDEF   be    a 

j  '  *—• 

polygon,  and  AB'  be  the  /    ^-  "\  B '       C ' 

straight  line.  /?^— ^,          \      A>± 

To  construct  on  A'B'  a 
polygon  similar  to  A-F. 

It  is  known  (from  224) 
that  two  polygons  are 
similar  when  they  are  composed  of  the  same  number  of  similar 


§240.]  RATIO  AND  PROPORTION.  99 

triangles  similarly  placed;  therefore,  divide  A-Fiuto  triangles 
by  drawing  the  diagonals  AC,  AD,  and  AE. 

It  is  known  (from  218)  that  triangles  are  similar  when  they 
are  equiangular. 

Therefore,  construct  (by  187)  on  A'B'  a  triangle  equiangular 
with  ABC,  say  A'B'C' ;  then  on  AC'  a  triangle  equiangular 
with  ACD,  say  A'C'D'.  Likewise  on  AD',  AD'E',  and  on 
AE1,  A'E'F'-,  then  will  A'-F'  be  the  polygon  required. 


BOOK  IV. 

AREAS   OF   POLYGONS. 


DEFINITIONS. 

241.  The  Area  of  a  surface  is  the  numerical  value  of  the 
ratio  of  this  surface  to  another  surface,  called  the   Unit  of 
Surface)  or  Superficial  Unit. 

242.  The  unit  of  surface  is  the  square  whose  side  is  some 
Unit  of  Length,  as  an  inch,  a  foot,  a  metre,  etc.,  and  the  area 
is  expressed  as  so  many  square  inches,  square  feet,  square 
metres,  etc. 

243.  Two   surfaces   are   equivalent   when   their   areas   are 
equal. 

244.  The  projection  of  a  point  upon  a  straight  line  is  the 
foot  of  the  perpendicular  let  fall  from  the  point  upon  the  line. 
Thus  A'  is  the  projection  of  A. 

A 


The  projection  of  a  limited  straight  line  upon  another 
straight  line,  is  the  portion  of  the  latter  included  between  the 
projection  of  the  terminal  points  of  the  former.  Thus  A'B' 
is  the  projection  of  AB  on  XX'. 

100 


BK.  IV.,  §  245.]  AREAS   OF  POLYGONS.  101 


PROPOSITION  I.     THEOREM. 

245.    Two  rectangles  *  having  equal  altitudes  are  to  each  other 
as  their  bases. 


D 


! 

1 
1 

C            U 

\ 
1 

| 

1 
1 

1 
I 

1 

| 

1 

| 

1 

i 

R         A 

i 

F 


O 


O 


Let  the  two  rectangles  be  AC  and  AF,  having  the  same 

altitude  AD. 

AECD     AB 


To  prove  that 


AEFD ~  AE 


CASE  I.    When  the  bases  are  commensurable. 

* 

Let  AO  be  a  common  measure  of  AB  and  AE,  and  upon 
application  it  is  found  to  be  contained  7  times  in  AB  and 
4  times  in  AE. 

At  each  point  of  division  along  AB  erect  perpendiculars, 
and  likewise  on  AE.  This  will  divide  the  first  rectangle  into 
7  rectangles  and  the  second  into  4. 

These  small  rectangles  are  equal,  since  having  all  parts  the 
same  they  can  be  applied  one  to  the  other  and  will  coincide 
throughout,  thus 

ABCD     7 


AEFD  ~4 


but 


Therefore  (by  28) 


AE     4 


ABCD  =  AB 
AEFD     AE 


*  By  rectangles  is  meant  the  area  of  the  rectangles. 


102 


PLANE  GEOMETRY. 


[BK.  IV. 


CASE  II.    When  AB  and  AE  are  incommensurable. 


D 


A 


1 
1 

1 

1 

1 

C 


B 


D 


A 


H 


K 


In  this  case  we  find  a  portion  of  AE,  say  AK,  which  is 
commensurable  with  AB,  erect  the  perpendicular  KH\  then 

(from  first  case), 

ABCD  =  AB 

AKHD     AK 

By  diminishing  the  common  measure  a  larger  portion  of  AE 
can  be  found  which  will  be  commensurable  with  AB,  but  the 
above  equality  of  ratios  will  exist. 

The  limit  of  AKHD  is  AEFD,  and  the  limit  of  AK  is  AE. 


Therefore  (by  172) 


ABCD  =  AB 
AEFD  ~  AE' 


Q.E.D. 


246.  COR.     Since  either  side  of  a  rectangle  may  be  taken 
as  the  base,  it  follows  that 

Two  rectangles  having  equal  bases  are  to  each  other  as  their 
altitudes. 

PROPOSITION  II.     THEOREM. 

247.  Any  tivo  rectangles  are  to  each  other  as  the  products  of 
their  bases  by  their  altitudes. 

NOTE.     By  the  product  of  two  lines  is  to  be  understood  the  product  of 
their  numerical  measures  when  referred  to  a  common  unit  (§  242). 


a 


a' 


B 


b' 


a 


b' 


§  248.] 


AREAS   OF  POLYGONS. 


103 


Let  A  and  B  be  any  two  rectangles  having  the  altitudes 
a  and  a',  and  the  bases  b  and  &',  respectively. 

B  =  a'  xb' 
A      a  x  b 


To  prove  that 


Construct  a  rectangle  (7,  with  a  base  equal  to  the  base  of  B 
and  altitude  equal  to  that  of  A. 

Then  (by  246),  comparing  B  and  C, 

B  =  a[ 
C      a' 

Likewise  (by  245),  comparing  C  and  A, 


_ 
A     b 

Multiplying  these  proportions  (by  207), 

B  x  C      a'  x  b' 


or 


C  xA 

B 
A 


a  xb 
a'  xb' 

< 

a  x  b 


Q.E.D. 


PROPOSITION  III.     THEOREM. 

248.    The  area  of  a  rectangle  is  equal  to  the  product  of  its  base 
and  altitude. 


R 


Let  R  be  the  rectangle,  b  the  base,  and  h  the  altitude  j  and 
let  S  be  a  square  whose  side  is  the  linear  unit. 
To  prove  the  area  of  R  =  h  x  b. 


104  PLANE  GEOMETRY.  [BK.  IV. 

It  is  known  (from  247)  that  two  rectangles  are  to  each  other 
as  the  products  of  their  bases  by  their  altitudes ;  therefore, 

R     h  X  b         7       ^ 

s=na'  =hxb> 

but  S  is  the  unit  of  area ; 

hence  R  =  h  x  b. 

249.  COR.     If  h  =  b,  then  R  =  b  x  b  =  tf. 

But  when  the  base  and  altitude  of  a  rectangle  are  equal,  the 
figure  (by  105)  is  a  square,  hence  the  area  of  a  square  is  equal 
to  the  square  of  one  of  its  sides. 

250.  SCHOLIUM.     The  statement  of  this  proposition  is  an 
abbreviation  of  the  following: 

The  number  of  units  of  area  in  a  rectangular  figure  is  equal  to 
the  product  of  the  number  of  linear  units  in  its  base  by  the  num- 
ber of  linear  units  in  its  altitude. 

PROPOSITION  IV.     THEOREM. 

251.  The  area  of  a  parallelogram  is  equal  to  the  product  of  its 
base  and  altitude. 

Let  ABCD  be  a  parallelogram,  and  EB  its  altitude. 

To  prove  that  the  area  of  F  Z)  EC 

ABCD  =  ABx  EB. 

Erect  the  perpendiculars  AF&nd  BE 
and  produce  CD  to  F,  forming  the  rec- 
tangle ABEF.  A 

In  the  right  triangles  ADF  and  BCE  the  sides  AD  and  BC 
are  (by  106)  equal,  and  JLFand  BE  are  (by  108)  equal;  there- 
fore, the  triangles  are  equal. 

If  from  the  entire  figure  ABCF  the  triangle  ADF  be  sub- 
tracted, the  parallelogram  ABCD  remains ;  and  if  from  the 


§254.]  AREAS  OF  POLYGONS.  105 

same  figure  the  equal  triangle  BEC  be  subtracted,  the  rectangle 
ABEF  remains. 

Therefore  (by  28) 

ABCD  =  ABEF. 

But  (by  248)  ABEF  =  AB  x  EB. 

Hence  ABCD  =  AB  x  EB.  Q.E.D. 

252.  COR.  1.     Parallelograms  having  equal  bases  and  equal 
altitudes  are  equivalent,  because  they  are  all  equivalent  to  the 
same  rectangle. 

253.  COR.  2.     Any  two  parallelograms  are  to  each  other  as 
the  products  of  their  bases  by  their  altitudes;  therefore,  parallelo- 
grams having  equal  bases  are  to  each  other  as  their  altitudes,  and 
parallelograms  of  equal  altitudes  are  to  each  other  as  their  bases. 

EXERCISES. 

1.  Show  that  the  diagonals  of  a  parallelogram  divide  it  into  four 
equivalent  triangles. 

2.  Show  that  the  area  of  a  rhombus  is  equal  to  one-half  the  product  of 
its  diagonals. 

PROPOSITION  V.     THEOREM. 

254.  The  area  of  a  triangle  is  equal  to  one-half  the  product  of 
its  base  and  altitude. 

A. 


Let  ABC  be  a  triangle,  having  its  altitude  equal  to  h  and  its 
base  equal  to  b. 

To  prove  that  area  ABC  =  ^  h  x  b. 


106  PLANE   GEOMETRY.  [BK.  IV. 

Draw  the  lines  AD  and  CD  parallel  to  BC  and  AB. 

Then  ABCD  is  a  parallelogram  having  its  altitude  equal  to  h 
and  its  base  equal  to  b. 

It  is  known  (from  107)  that  the  diagonal  of  a  parallelogram 
divides  it  into  two  equal  triangles  ;  therefore 

ABC  =  i  ABCD. 

But  (by  251)  ABCD  =  h  x  b. 

Therefore  ABC  =  \  h  x  b.  Q.E.D. 

255.  COR.  1.     Two  triangles  having  equal  bases  and  equal 
altitudes  are  equivalent. 

256.  COR.  2.     Two  triangles   having   equal  altitudes  are  to 
each  other  as  their  bases;  two  triangles  having  equal  bases  are  to 
each  other  as  their  altitudes;  and  any  two  triangles  are  to  each 
other  as  the  products  of  their  bases  by  their  altitudes. 

257.  COR.  3.     A  triangle  is  equivalent  to  one-half  of  a  paral- 
lelogram having  the  same  base  and  altitude. 

EXERCISES. 

1.  The  area  of  a  rectangle  is  6912  square  inches  and  its  base  is  2  yards. 
What  is  its  perimeter  in  feet  ? 

2.  If  the  base  and  altitude  of  a  triangle  are  18  and  12,  what  is  the 
length   of    the  side   of    an 

equivalent  square  ? 

3.  Show  that  the  area  of 
a  triangle  is  equal  to  one- 
half    the    product    of    its 
perimeter  by  the  radius  of 
the  inscribed  circle. 

SUGGESTION.  Join  the 
centre  with  each  vertex, 
and  find  the  area  of  OB  C, 
OB  A,  and  OAC.  C 


§260.]  AREAS   OF  POLYGONS.  107 


PROPOSITION  VI.     THEOREM. 

258.    The  area  of  a  trapezoid  is  equal  to  the  product  of  the 
half  sum  of  its  parallel  sides  by  its  altitude. 


Let  ABCD  be  a  trapezoid,  with  AB  and  CD  its  parallel 
sides  and  DH  the  altitude. 
To  prove  that  the  area  of 

ABCD  =  |  (AB  +  CD)  x  DH. 

Join  DB,  making  of  the  trapezoid  two  triangles. 
It  is  known  (from  254)  that  the  area  of 

ADB  =  ±ABx  DH, 

and  area  of  DCB  =  \  DC  x  DH. 

Hence  by  adding 

ADB  +  DCB  =  ±ABxDH  +  ±DCx  DH, 
or  ABCD  =  i  (AB  +  DC)  x  DH.  Q.E.D. 

259.  Since  (by  113)  the  median  line   FE  =  ±(AB  +  DC\ 
then  the  area  of  a  trapezoid  is  equal  to  the  product  of  the 
median  joining  the  middle  points   of   the   non-parallel   sides 
by  the  altitude. 

.-.  area  ABCD  =  FE  x  DH. 

260.  Occasionally  the  area  of  an  irregular  polygon  is  found 
by  dividing  the  figure  into  trapezoids  and  triangles,  and  find- 
ing the  area  of  each  and  taking  their  sum. 


108 


PLANE  GEOMETRY. 


[BK.  IV. 


EXERCISE. 

1.    In   a  trapezoid    the    straight 
lines,  drawn  from  the  middle  point 
of   one  of  the  non-parallel  sides  to         ,_, 
the  ends  of  the  opposite  side,  form 
with  that  side  a  triangle   equal  to    ~ 
half  the  trapezoid. 

SUGGESTION.     Compare  area  of  ABE,  BE F  and  FEC,  EDC. 


PROPOSITION  VII.     THEOREM. 

261.  The  areas  of  two  triangles  having  an  angle  of  the  one 
equal  to  an  angle  of  the  other,  are  to  each  other  as  the  products  of 
the  sides  including  the  equal  angles.  * 

Let  ABC  and  ADE  be  two  triangles, 
having  /.  A  common.  j) 

ABC      AB  x  AC 


Join  BE,  then  the  two  triangles  ABC 
and  ABE  having  their  bases  in  the  same  &  C 

line  AC  and  their  vertices  in  the  same  point  EB  will  have  the 
same  altitude,  hence  (by  256) 

ABC  =  AC 
ABE     AE 

Likewise  the  triangles  ABE  and  ADE  having  their  bases  in 
the  same  line  (AB),  and  their  vertices  in  the  same  point  (E), 
will  have  the  same  altitude,  hence 

ABE  =  AB 
ADE  ~  AD 

Multiplying  these  ratios  (by  207), 

ABC  x  ABE     AC  x  AB 


or 


ABE  x  ADE  "  AE  x  AD 

ABC  =  AC  xAB 
ADE~  AE  xAD 


Q.E.D. 


§264.]  AREAS   OF  POLYGONS.  109 

PROPOSITION  VIII.     THEOREM. 

262.    Two  similar  triangles  are  to  each  other  as  the  squares  of 
their  homologous  sides. 

C 


Let  AC  and  A'C1  be  homologous  sides  of  the  similar  tri- 
angles ABC  and  A'B'C'. 

ABC       AC* 
To  prove  that  -  =  -      -• 

A'B'C'     ^C'2 

The  triangles  being  similar,  they  are  (by  223)  equiangular, 

therefore  (by  261) 

ABC        AC  x  BC 


A'B'C'     AC'  x  B'C' 

But  as  the  triangles  are  similar,  we  have  (by  222) 

BC       AC 


proportion,  it  gives 


B'C'   'A'C1 

i.tio.  for 

B'C1 


BC 

Therefore,  substituting  this  equal  ratio,  for  — —  in  the  above 


ABC        AC  x  AC       AC2 

-•  Q.E.D. 


A'B'C'     AC'  x  AC      A^c12 

263.  SCHOLIUM.     Two  similar  triangles  are  to  each  other  as 
the  squares  of  any  two  homologous  lines. 

264.  COR.    Tivo  similar  polygons  are   to   each  other  as   the 
squares  of  their  homologous  sides. 

Similar  polygons  (by  224)  can  be  divided   into  the   same 
number   of   similar   triangles,  and  (by  209)  the    sum  of   the 


110 


PLANE  GEOMETRY. 


[BK.  IV. 


triangles  of  one  polygon  will  be  to  the  sum  of  the  triangles 
of  the  other  as  any  one  triangle  of  the  former  is  to  a  corre- 
sponding triangle  of  the  latter. 

But  these  triangles  are  to  each  other  (by  262)  as  the  squares 
of  their  homologous  sides. 

Therefore  the  sums  of  the  triangles  or  polygons  are  to  each 
other  as  the  squares  of  their  homologous  sides. 


PROPOSITION  IX.     THE  PYTHAGOREAN  THEOREM.* 

265.  In  any  right  triangle  the  square  described  upon  the 
hypotenuse  is  equivalent  to  the  sum  of  the  square  described  upon 
the  other  two  sides. 

First  Method. 


A  D  B 

Let  ABC  be  a  right  triangle. 

To  prove  that  the  square  described  upon  the  hypotenuse  AB 
is  equivalent  to  the  sum  of  the  squares  described  upon  the 
sides  AC  and  BC. 

Draw  CD  perpendicular  to  AB. 

Then  (by  227)  AC2  =  AB  x  AD, 

and 

Adding,  we  have 


BC2  =  ABx  BD. 


AC  +  BC  =ABx  (AD  +  BD) 
=  AB  x  AB 

=  AB2. 

*  This  proposition  is  called  the  Pythagorean  Proposition  because  it  is 
said  to  have  been  first  given  by  Pythagoras  (born  about  600  B.C.). 


§  265.] 


AREAS  OF  POLYGONS. 


Ill 


But  AC  ,  BC~,  and  AE"  are  the  areas  of  the  squares  de- 
scribed upon  the  sides  AC,  BC,  and  AB  (by  249). 

Hence  the  square  described  upon  AB  is  equivalent  to  the 
sum  of  the  squares  described  upon  AC  and  BC. 

Second   Method. 
G 


L  E 


Construct  upon  AC,  the  square  ACGF,  upon  CB,  CBKff, 
and  upon  AB,  ABED. 

Draw  CL  perpendicular  to  DE,  and  join  FB  and  CD. 

The  triangle  FAB  is  one-half  the  square  ACGF,  having  the 
same  base  AFand  the  same  altitude  AC. 

The  triangle  DAC  —  1 ADLN,  having  the  same  base  AD 
and  the  same  altitude  AN. 

The  triangles  FAB  and  CAD  are  equal,  having  the  sides 
FA  =  AC,  being  sides  of  the  same  square,  and  for  the  same 
reason  AB  =  AD.  The  included  angles  FAB  =  Z.  CAD,  both 
being  equal  to  a  right  angle  plus  the  common  angle  CAB. 
These  two  triangles  are  therefore  (by  86)  equal. 

As  these  triangles  are  equal, 


or 


%  ACGF =|  ADLN, 
ACGF '=  ADLN. 


•v 


112 


PLANE  GEOMETRY. 


[BK.  IV. 


In  a  similar  manner  by  joining  AK  and  CE,  it  can  be  shown 
that 

CHKB  =  NLEB, 
or  by  addition, 

ACGF+  CHKB  =  ADLN+  NLEB 

=  ABED, 

or  AB2  =  AC2  +  CB2. 

266.   COR.  1.     From  the  last  equation,  by  transposition, 

AC2  =  A$  -  CB2, 
and 


CB  =  AB"  -  AC, 


or 
and 


CB  = 


AC. 


EXERCISES. 

1.  Show  that  the  diagonal  and  side  of  a  square  are  incommensurable. 

2.  Find  the  length  of  the  diagonal  of  a  rectangle  whose  area  is  96  and 
whose  altitude  is  8. 

3.  Show  that  if  similar  polygons  be  similarly  drawn  on  the  sides  of  a 
right  triangle,  the  polygon  on  the  hypotenuse  is  equal  to  the  sum  of  the 
polygons  on  the  other  sides. 

PROPOSITION  X.     THEOREM. 

267.  In  any  triangle,  the  square  on  the  side  opposite  an  acute 
angle  is  equivalent  to  the  sum  of  the  squares  of  the  other  two  sides 
diminished  by  twice  the  product  of  one  of  those  sides  and  the  pro- 
jection of  the  other  upon  that  side. 


Fur.  1. 


Fi<r.  2. 


§268.]  AEEAS   OF  POLYGONS.  113 

Let  C  be  an  acute  angle  of  the  triangle  ABC,  and  DC  the 
projection  of  AC  upon  BC. 
To  prove  that 

AS  =  BC2  -J-  AC2  -2BCx  DC. 

There  will  be  two  possible  cases ;  in  the  first  the  projection 
of  A  will  be  within  the  base  of  the  triangle  (Fig.  1) ;  in  the 
second  it  will  be  on  the  base  produced  (Fig.  2). 

In  the  first  case, 

DB  =  BC-DC-, 
in  the  second, 

DB=DC-  BC. 

Squaring  in  either  case, 

DB2  =  DC2  +  BC2  -2DCx  BC 


Add  AD  to  both  sides  of  the  equation  ;  then 


AD2  +  DB  =  AD  +  DC  +  BC  -2DCx  BC. 
But  (by  265), 


AD  +  DB"  =  AB",  and  AD'  +  DC  =  AC  ; 


therefore  AB"  =  AC~  +  BC  -2DCx  BC. 

268.   COR.     Another  proof,  using  algebraic  processes,  is : 


By  (265)  AB~  =  AD'  +  BD. 

Bat  (by  266)  AD2  =  AC2  -  DC*. 

Also  BD  =  DC-  BC,  (Fig.  2), 

which  gives,  by  substitution, 

AB2  =  AC2  -  DC2  +  (DC  -  BC)2 


+  DC-2DCxBC+  BC' ; 

or  by  cancellation, 


AB  =  AC  +  BC"  -  2  DC  x  BC.  Q.E.D. 


114  PLANE  GEOMETRY.  [Bit.  IV. 


PROPOSITION  XI.     THEOREM. 

269.  In  an  obtuse-angled  triangle,  the  square  of  the  side  oppo- 
site the  obtuse  angle  is  equal  to  the  sum  of  the  squares  of  the  other 
two  sides  increased  by  tivice  the  product  of  one  of  these  sides  by 
the  projection  of  the  other  side  upon  it. 


C  B         D 

Let  ABC  be  a  triangle  with  obtuse  angle  ABC. 
To  prove  that 


I  y^V  •* 


AC  =  ABT  +  EC  +2BCx  BD. 
BC. 


Squaring,  CD  =  BD  +  BC*  +  2BCx  BD. 

Add  AD"  to  both  sides  of  the  equation, 


=  AD  +  BD'  +  BC  +2BCx  BD. 
But  (by  265) 

Aff  +  CZ?  =  AC?,  and  AD2  +  BD2  =  AB2. 

Making  these  substitutions, 

AC2  =  AB2  +  BC2  +  2BCx  BD.  Q.E.D. 

270.  COR.  From  the  three  preceding  theorems,  it  follows  that 
the  square  of  the  side  of  a  triangle  is  less  than,  equal  to,  or  greater 
than,  the  sum  of  the  squares  of  the  other  two  sides,  according  as  the 
angle  opposite  this  side  is  acute,  right,  or  obtuse. 


§271.] 


AREAS   OF  POLYGONS. 


115 


EXERCISES. 

1.  Prove  the  above  by  the  method  of  268. 

2.  Show  that  the  sum  of  the  squares  on  the  diagonals  of  a  parallelo- 
gram is  equal  to  the  sum  of  the  squares  on  the  four  sides. 

SUGGESTION.    Apply  263  and  264. 

3.  The   sum   of    the   squares 
upon  the  diagonals  of  a  trape- 
zoid  is  equal  to  the  sum  of  the 
squares    upon   the    non-parallel 
sides  plus  twice  the  rectangle  of 
the  parallel  sides. 

CONCLUSION.     AC2  +  BD'2  =  AD'2  +  BC2  +  2ABx  CD. 

SUGGESTION.     Take  the  triangles  ACS  and  B CD,  and  apply  207  and 
269. 

PROPOSITION  XII.     PROBLEM. 

271.    To  find  the  area  of  a  triangle  when  its  three  sides  are 
given. 


c 


Let  a,  6,  and  c  denote  the  three  sides  of  the  triangle  ABC, 
and  draw  AD  perpendicular  to  BC. 

Then  if  C  is  an  acute  angle,  we  have  (by  267), 

c2  =  a2  +  b2  -  2  a  x  CD, 


or 


2  a 


116  PLANE  GEOMETRY.  [BK.  IV. 


Now  (by  266)      AD2  =  AC2  - 


4  a2 

=  4  a262  -  (a2  +  b2  -  c2)2 

~~  4  a2 

The  second  member  of  this  equation  is  the  difference  of  two 
squares,  and  hence  can  be  factored ;  that  is, 

FT?      [2  ab  +  (a2  +  62  -  c2)]  [2  a&  -  (a2  +  b2  -  c2)] 

^ll>  — T2 

4  a2 

=  (2  afr  +  a2  +  b2  -  c2)(2  a6  -  a2  -  62  +  c2) 
~  4  a2 

=  [(a  +  b)2  -  c2]  [c2  -  (a  -  6)2] 
4  a2 

=  [(a  +  b  -  c)  (a  +  b  +  c)]  [c  -(a  -  6)1  [c  +a  - 

4  a2 

(a  +  &  --  c)  (a  +  b  +  c)  (c  --  a  +  6)  (c  +  a  -  - 
~  4  a2 

Let  a-|-fr-f-c  =  2s. 

Subtract  2  c  =  2  c, 

then  a  +  6  —  c  —  2s  —  2c  =  2(s  —  c). 

Similarly,  a  -f  c  —  &  =  2  (s  —  5), 

and  —  a  +  c  -\-b  =  2(s  --  a), 

Substituting  these  values  in  the  above  equation, 

__  2s'2(s-a)2(s-b)2(s-c) 

—  _ 

4  a2 


a2 


2  

/~^V                                                                                                   /I      /    J            —           _  -y    /    o    /    O                      /"/    \/O                      Al    I/O     —            /"*    I 

\_/  A.                                                                 *\  _M  J    '      *        '  \'    OlO    ""•"    (.t/llO               ly  /  \  O          ~    v  / » 

d 


§271.]  AREAS   OF  POLYGONS.  Ill 

But  (by  254),    Area  of  ABC  =  1  BC  x  AD 


2     _ 
=  la-  Vs(s  -  a)(s  -  b)(s  -  c) 


=  V«  (s  —  «)(s  ~~  fr)(s  ~~ 
In  which  s  is  one-half  of  the  perimeter. 


EXERCISES. 

1.  If  the  sides  of  a  triangle  are  13,  14,  15,  find  the  area. 

2.  In  the  above,  find  the  radius  of  the  inscribed  circle  (see  257,  Ex.  3). 

3.  The  area  of  a  rhombus  is  24  and  its  side  is  5  ;  find  the  lengths  of 
the  diagonals. 

4.  If  the  sides  of  an  isosceles  triangle  are  a,  a,  and  6,  show  that  its 

area  is  -  V4  a2  —  b'2. 
4 

5.  If  from  any  point  on  the  diagonal 
of  a  parallelogram  lines  be  drawn  to 
the  opposite  angles,  the  parallelogram 
will  be  divided  into  two  pairs  of  equal 
triangles. 

Area  OA  D  =  area  OAB. 
Area  OCD  =  area  OCB. 

SUGGESTION.      OAD  =  \  OA  x  DF,   OA  B  =  ~  OA  x  BE  ;  hence  show 
that  D  F  =  EB. 

6.    Show  that  two  quadrilaterals  are  equivalent  when  they  have  the 
following  parts  of  the  one  respectively  equal  to  the  corresponding  parts  in 

the  other  : 

I.    Four  sides  and  one  diagonal. 

II.    Four  sides  and  one  angle. 

III.  Two  adjacent  sides  and  three  angles. 

IV.  Three  sides  and  the  two  included  angles. 

7.  Construct  a  square  having  a  given  diagonal. 

8.  Show  that  the  sum  of  the  squares  of  the  sides  of  a  triangle  is  equal 
to  double  the  square  of  the  bisector  of  the  base  together  with  double  the 
square  of  half  the  base. 


118 


PLANE  GEOMETRY. 


[Bit.  IV, 


PROBLEMS   IN  CONSTRUCTION. 
PROPOSITION  XIII.     PROBLEM. 

272.    To  construct  a  square  equivalent  to  the  sum  of  tivo  given 
squares. 


c 

M 

N 

X 


Let  M  and  N  be  the  given  squares. 

To  construct  a  square  equivalent  to  their  sum. 

It  is  known  (from  265)  that  the  square  upon  the  hypotenuse 
is  equivalent  to  the  sum  of  the  squares  on  the  other  two  sides. 

Therefore  construct  a  right  triangle  whose  base  will  be  equal 
to  a  side  of  M,  say  AB,  and  whose  altitude  will  be  equal  to  a 
side  of  N9  say  AC,  then  the  hypotenuse,  say  BC,  will  be  a  side 
of  the  square  required. 

EXERCISES. 

1.  To  construct  a  square  equivalent  to  the  sum  of  any  number  of 
squares. 

2.  To  construct  a  square  equivalent  to  the  difference  of  two  given 
squares. 

PROPOSITION  XIV.     PROBLEM. 
273.    To  construct  a  square  equivalent  to  a  given  parallelogram. 


H 


C 


A     E 


§274.] 


AREAS  OF  POLYGONS. 


119 


Let  ABCD  be  the  given  parallelogram. 

To  construct  a  square  equivalent  to  ABCD. 

It  is  known  (from  251)  that  the  area  of  the  parallelogram 
ABCD  =  AB  x  DE,  therefore  any  square  to  be  equivalent  to 
ABCD  must  have  such  a  side  that  its  square  must  be  equal  to 
AB  x  DE. 

It  is  known  (from  200)  that  when  the  square  of  one  quantity 
is  equal  to  the  product  of  two  other  quantities  the  former  is 
said  to  be  a  mean  proportional  to  the  other  two. 

Hence  find  (by  238)  a  mean  proportional  to  AB  and  DE,  say 
FG,  then  the  square  on  FG  will  be  the  required  square  equiva- 
lent to  ABCD. 

EXERCISES. 

1.  To  construct  a  square  equivalent  to  a  given  triangle. 

2.  To  construct  a  square  equivalent  to  the  sum  of  two  given  triangles. 


PROPOSITION  XV.     PROBLEM. 

274.    Upon  a  given  straight  line,  to  construct  a  rectangle  equiva- 
lent to  a  given  rectangle. 

H  G 


D 


B 


E 


Let  ABCD  be  the  given  rectangle,  and  EF  the  given  line. 

To  construct  upon  EF  as  a  base  a  rectangle  equivalent  to 
ABCD. 

The  area  of  the  given  rectangle  is  AB  x  DA,  therefore  if 
EF  is  the  base  of  the  required  rectangle,  its  altitude  must  be 
such  a  value  that  when  multiplied  by  EF  the  product  will  be 
equal  to  AB  x  AD ;  that  is,  a  fourth  proportion?!  to  EF,  AB, 
and  DA. 


120 


PLANE  GEOMETRY. 


[BK.  IV. 


Therefore  find  (by  237)  a  fourth  proportional  to  EF,  AB, 
DA  j  suppose  it  is  HE,  that  is 

EF:AB  =  DA:HE,   or 
EFxHE  =  ABx  DA, 
or  area  HEFG  =  area  ABCD. 


PROPOSITION  XVI.     PROBLEM. 

275.    To  construct  a  rectangle   equivalent  to  a  given  square, 
having  the  sum  of  its  base  and  altitude  equal  to  a  given  line. 


C      D 


\ 


\ 


1 


N 


Let  M  be  the  given  square  and  AB  the  given  line. 

To  construct  a  rectangle  equivalent  to  M,  having  the  sum 
of  its  base  and  altitude  equal  to  AB. 

It  is  known  (from  228)  that  the  perpendicular  let  fall  from 
any  point  in  the  circumference  upon  the  diameter  is  a  mean 
proportional  between  the  segments  into  which  it  divides  the 
diameter. 

Hence  we  take  AB  as  the  diameter  (183)  and  find  a  point 
on  the  circumference  which  is  as  far  from  the  diameter  as  the 
side  of  the  square. 

To  do  this,  erect  at  A  (by  184,  Ex.  1)  a  perpendicular  to  AB. 
equal  to  a  side  of  M,  say  AC,  through  C,  draw  a  line  parallel  to 
AB  (by  189),  say  CF;  then  where  CF  intersects  the  circumfer- 
ence, say  D,  let  fall  the  perpendicular  DE. 

We  know  (by  228) 


but 


DE2  =  M. 


§276.]  AREAS   OF  POLYGONS.  121 

Construct  a  rectangle  JY"  whose  base  and  altitude  are  EB 

and  AE,  then 

N=AE  xEB, 

or 


PROPOSITION  XVII.     PROBLEM. 

276.    To  construct  a  square  having  a  given  ratio  to  a  given 
square. 

C 


/ 


m 


71  /  /  ^     ' 

«fr  I  /  v  i 

K.  j 

AT~m~  "D      ~^        B 

Let  M  be  the  given  square,  and  let  the  given  ratio  be  that 
of  the  lines  m  and  n. 

To  construct  a  square  which  shall  have  to  M  the  ratio  m  :  n. 

It  is  known  (from  228)  that  the  perpendicular  let  fall  from 
any  point  in  the  circumference  upon  the  diameter  divides  it 
into  segments  which  have  the  same  ratio  as  the  squares  of  the 
chords  drawn  from  the  same  point  to  the  two  extremities  of 
the  same  diameter. 

Hence  we  lay  off  on  a  straight  line  DA  =  m,  and  DB  =  n, 
and  on  AB  erect  (by  183)  a  semicircumference. 

At  D  erect  (by  182)  the  perpendicular  DC,  and  join  CA 
and  CB.  

Then  (by  228),  CA2 :  CB2  =  m  :  n. 

But  neither  CA  nor  CB  is  a  side  of  the  square  M,  that  is, 
we  must  take  a  part  of  CA,  say  CE,  that  is  equal  to  a  side  of 
M,  and  find  some  quantity  that  has  the  same  ratio  to  CE  that 
CB  has  to  CA. 

It  is  known  (from  213)  that  a  line  drawn  through  E  par- 
allel to  AB  will  divide  CB  into  parts  having  the  same  ratio  as 
the  parts  into  which  E  divides  CA. 


122 


PLANE  GEOMETRY. 


[BK.  IV. 


Therefore  draw  (by  189)  EF  parallel  to  AB ; 
then  CA:CB=CE:  CF, 

or  (by  208)  CA2 :  CB*  =  CE* : 

but 

therefore 


CA  :  CB  =  m:n, 
CE2:CF2  =  m :  n, 


or  CF  is  the  side  of  the  square  required. 

COR.     If  a  side  of  M  is  greater  than  CA,  extend  CA  and 
CB,  and  proceed  in  the  same  manner. 


C 


m 

.-''/' 

/  / 

N\    "\x 

^v                                     S. 

n 

/L/ 

,          r 

:  / 
t/ 

X^     \ 

"*""""""""     ^-v           \ 

\  \ 

N 

A.    m    i 

)       n       B 

EXERCISES. 

1.  To  construct  a  square  equivalent  to  the  sum  of  a  given  triangle  and 
a  given  parallelogram. 

2.  To  construct  an  isosceles  triangle  equivalent  to  a  given  triangle,  its 
altitude  being  given. 

PROPOSITION  XVIII.     PROBLEM. 

277.    To  construct  a  triangle  equivalent  to  a  given  polygon. 

Let  ABCDE  be  the  given  polygon. 

To  construct  a  triangle  equivalent 
to  ABCDE. 

If   it   is  possible   to  construct   one    E 
polygon    equivalent    to    another    but       / 
with  one  side  less,  then  a  continuation 
of  this  operation  would  eventually  re-     Q 
suit  in  a  triangle. 

One  side  of  the  polygon,  say  AB,  can  be  extended  without 
increasing  the  number  of  sides,  then  draw  DA,  in  the  effort  to 


§277.]  AREAS   OF  POLYGONS.  123 

find  upon  DA  and  AB  produced  a  triangle  equivalent  to  DEA 
introducing  one  line  in  the  place  of  two,  that  is  DE  and  EA. 

If  DA  is  regarded  as  the  base,  then  the  required  triangle 
must  have  (by  255)  an  altitude  equal  to  the  distance  from  E 
to  DA ;  and  since  (by  60)  parallel  lines  are  everywhere  equally 
distant,  the  vertex  of  the  required  triangle  must  lie  on  the 
parallel  to  DA  drawn  through  E.  Again,  if  AB  produced  is 
to  be  a  side  of  the  triangle,  the  vertex  must  also  be  on  AB 
produced  or  at  G. 

Draw  DG,  and  the  triangle  DGA  will  be  the  equivalent  of 
DEA. 

Add  to  DABC  the  triangle  DEA,  and  we  have  the  original 
polygon;  add  to  the  same  figure  the  equal  triangle  DGA,  and 
we  have  the  polygon  DGBC;  therefore  DGBC  =  DEABC, 
and  has  one  side  less. 

Draw  OF  parallel  to  DB  and  draw  DF-,  then  the  triangle 
DGF  will  be  equivalent  to  the  polygon  ABODE. 

EXERCISES. 

1.  To  draw  a  square  equivalent  to  a  given  polygon. 

2.  To  construct  a  square  equal  to  two  given  polygons. 

3.  Two  similar  polygons  being  given,  to  construct  a  similar  polygon 
equal  to  their  sum. 

SUGGESTION.     See  240  and  272. 

4.  On  a  given  straight  line 
construct  a  triangle  equal  to  a 
given  triangle    and    having  its 

vertex  on  a  given  straight  line     /£~        jj      ~\> 
not  parallel  to  the  base. 

SUGGESTION.  Find  (by  237)  a  fourth  proportional  to  EF,  AB,  and 
|  CD,  and  it  will  be  the  required  altitude  ;  then  see  196,  Ex.  2. 

5.  When  is  the  last  problem  impossible  ? 


BOOK  V. 

REGULAR   POLYGONS   AND   CIRCLES. 

278.  A  Regular  Polygon  is  a  polygon  which  is  equilateral 
and  equiangular. 

PROPOSITION  I.     THEOREM. 

279.  A  circle  may  be  circumscribed  about,  or  inscribed  within, 
any  regular  polygon. 


Let  ABODE  be  a  regular  polygon. 

1.   To  prove  that  a  circle  may  be  circumscribed  about  it. 

Let  A,  B,  and  C  be  any  three  vertices,  and  through  them 
pass  (by  194)  a  circle  ;  let  its  centre  be  at  0.  Join  OA,  OB, 
OC,  OD,  and  OE. 

Since  the  polygon  is  equiangular, 

Z  ABC  =  Z  BCD, 

and  since  OB  =  OC,     Z  OBC  =  Z  OCB. 
Subtracting  these  equal  angles, 

Z  ABC-  Z  OBC=  Z  BCD  -  Z  OCB, 


01 


124 


BK.  V.,  fc  -285.]      REGULAR   POLYGONS   AND   CIRCLES.      125 

therefore  the  triangles  OCD  and  ABO  have  OC  =  OB  being 
radii,  AB  =  CD  sides  of  the  regular  polygon,  and  /.  OBA 
=  Z  OCD.  They  are  therefore  equal,  and  OD  =  OA. 

Hence  the  circle  passing  through  A,  B,  and  C,  also  passes 
through  D. 

In  the  same  manner  it  can  be  shown  to  pass  through  E. 

2.    To  prove  that  a  circle  may  be  inscribed  in  ABCDE. 

Since  AB,  BC,  CD,  DE,  and  EA  are  equal  chords,  they  are 
(by  148)  equally  distant  from  the  centre  0. 

Hence  if  a  circle  be  described  with  0  as  a  centre,  and  a 
radius  equal  to  the  perpendicular  distance  from  0  to  one  of 
the  sides,  the  circumference  will  touch  all  the  sides  of  the 
polygon.  Q.E.D. 

280.  The  Centre  of  a  regular  polygon  is  the  common  centre 
0  of  the  circumscribed  and  inscribed  circles. 

281.  The  Radius  of  a  regular  polygon  is  the  radius  OA  of 
the  circumscribed  circle. 

282.  The  Apothem  of  a  regular  polygon  is  the  radius  OF 
of  the  inscribed  circle. 

283.  The  Angle  at  the  centre  is  the  angle  included  by  the 
radii  drawn  to  the  extremities  of  any  side. 

284.  COB.  1.     Each  angle  at  the  centre  of  a  regular  polygon  is 
equal  to  four  right  angles  divided  by  the  number  of  sides  of  the 
polygon. 

Since  the  triangles  OAB,  OBC,  OCD,  etc.,  are  equal,  the 
angles  AOB,  BOC,  COD,  etc.,  are  equal. 

Therefore  each  angle  is  equal  to  four  right  angles  (by  51) 
divided  by  the  number  of  sides. 

285.  COR.  2.     If  a  regular  inscribed  polygon  is  given,  the 
tangents  at  the  vertices  of  the  given  polygon  form  a  regular 
circumscribed  polygon  of  the  same  number  of  sides. 


126 


PLANE   GEOMETRY. 


[BK.  V- 


286.  COR.  3.     If  a  regular  inscribed  polygon  ABCD  •••  is 
given,  the  tangents  at  the  middle  points  M,  N,  P,  etc.,  of  the 
arcs  AB,  BC,   CD,  etc.,  form  a  regular  , 
circumscribed  polygon  whose   sides  are 

parallel  to  those  of  the  inscribed  polygon, 
and  whose  vertices  A  ,  73',  C',  etc.,  lie  on 
the  radii  OAA,  OBB'  prolonged,  etc. 

For  the  sides  AB,  A'B'  are  parallel, 
being  perpendicular  to  OM. 

Since  B'M=  B'N  (by  234),  the  right 
triangles  MOB1  and  NOB1  are  (by  92)  equal,  hence  the  point 
B  is  on  the  bisector  OB  of  the  angle  MON. 

Likewise  C  and  C",  D  and  D',  are  on  the  same  line. 

287.  COR.  4.     If  the  chords  AM,  MB,  BN,  etc.,  be  drawn, 
the  chords  form  a  regular  inscribed  polygon  of  double  the 
number  of  sides  of  ABCD  •••. 

288.  COR.  5.     If  through  the  points  A,  B,  C,  etc.,  tangents 
are  drawn  intersecting  the  tangents  A'B',  B'C',  etc.,  a  regular 
circumscribed  polygon  is  formed  of  double  the  number  of  sides 
ofA'B'C'D'—. 


289.  COR.  6,  If  the  circumference  of  a  circle  is  divided  into 
any  number  of  equal  arcs,  their  chords  form  a  regular  polygon 
inscribed  in  the  circle. 

Since  (by  145)  equal  arcs  are  subtended  by  equal  chords,  if 
tiie  arcs  are  equal  the  chords  will  be  equal. 

And  (by  175)  each  angle  will  be  measured  by  one-half  of 
the  circumference  excepting  the  two  arcs  subtended  by  the 
two  equal  chords  forming  the  sides  of  the  angle,  hence  each 
angle  will  have  the  same  measure. 

Therefore  the  polygon  will  be  equiangular  and  equilateral; 
and  hence  regular, 


§290.]  REGULAR   POLYGONS  AND   CIRCLES.  127 

EXERCISES. 

1.  Show  that  the  interior  angle  of  a  regular  polygon  is  the  supplement 
of  the  angle  at  the  centre 

2.  Show  that  the  radius  drawn  to  any  vertex  of  a  regular  polygon 
bisects  the  angle  at  that  vertex. 

3.  Show  that  if  the  circumference  of  a  circle  be  divided  into  any  num- 
ber of  equal  arcs,  the  tangents  at  the  points  of  division  form  a  regular 
polygon  circumscribed  about  the  circle. 


PROPOSITION  II.     THEOREM. 

290.   Regular   polygons   of   the   same   number   of  sides  are 
similar. 

Let  ABCDEFamlAB'C'D'E'F'  be  two  regular  polygons  of 

the  same  number  of  sides. 

A.    G    R 
To  prove  that  they  are 

similar. 


The  sum  of  the  angles 
of  the  one  polygon  is  (by 
123)  equal  to  the  sum  of 
the  angles  of  the  other; 
then  since  the  number  of  angles  are  the  same  in  both,  each 
angle  of  the  one  will  be  equal  to  the  corresponding  angle  of 
the  other.  Therefore  they  are  mutually  equiangular. 

The  polygons  being  regular, 

AB  =  BC  =  CD,  etc., 

and  A'B'  =  B'C  =  C'D',  etc. 

Dividing  the  former  equation  by  the  latter,  we  have 


_=^= 
A'B'"B'C'~      ''' 


Therefore  the  polygons  are  (by  223)  similar.  Q.E.D. 


128 


PLANE   GEOMETRY. 


[BK.  V. 


291.   COR.  1.     Taking  the  above  proportion  by  composition 
(204), 

BC+CD  +  etc,          AB 


or 


= 
A'B'  +  B'C'  +  C'D'  +  etc.  =  :  A'B1' 


Perimeter  of  AB  -  F        AB 


Perimeter  of  A'B'    -F'~~  A'B' : 
that  is, 

The  perimeters  of  regular  polygons  of  the  same  number  of  sides 
are  to  each  other  as  any  two  homologous  sides  or  lines. 


PROPOSITION  III.     THEOREM. 

292.    The  area  of  a  regular  polygon  is  equal  to  one-half  the 
product  of  its  perimeter  and  apothem. 


F      B 


Let  r  denote  the  apothem  OF,  and  P  the  perimeter  of  the 
regular  polygon  ABODE. 

To  prove  that  area  ABODE  =  J  P  x  r. 

By  drawing  the  radii  OA,  OB,  GO,  etc.,  the  polygon  may  be 
divided  into  a  series  of  triangles,  OAB,  OBC,  etc.,  whose  com- 
mon altitude  is  r. 

Then  (by  254)       area  OAB  =  ±ABxr, 

area  OBC  =  4-  BC  X  r.  etc. 

^j 

Adding,  we  have 

area  OAB  +  area  OBC  +  etc.  =  \  (AB  +  BC  +  etc.)  x  r. 
That  is,  area  ABODE  =  1 P  x  r.  Q.E.D. 


§295.]  EEGULAE   POLYGONS   AND   CIRCLES.  129 

EXERCISE. 

1.    The  apothem  of  a  regular  pentagon  is  6,  and  a  side  is  4  ;  find  the 
perimeter  and  area  of  a  regular  pentagon  whose  apothein  is  8. 

SUGGESTION.     See  291. 

PROPOSITION  IV.     PROBLEM. 
293.    To  inscribe  a  square  in  a  given  circle. 


D 

Let  0  be  the  centre  of  the  given  circle. 

To  inscribe  a  square  within  it. 

Since  the  inscribed  figure  is  to  be  a  regular  quadrilateral, 
each  angle  at  the  centre  will  be  one-fourth  of  four  right  angles, 
or  one  right  angle. 

Therefore  draw  any  diameter,  say  AC,  and  another  perpen- 
dicular thereto,  say  BD,  and  join  the  ends  of  these  diameters, 
and  the  figure  inscribed  will  be  the  square  required. 

294.  COR.  1.     If  tangents  be  drawn  to  the  circle  at   the 
points  A,  B,  C,  D,  the  figure  so  formed  will  be  a  circum- 
scribed square. 

295.  COR.  2.     To  inscribe  and  circumscribe   regular  poly- 
gons of  8  sides,  bisect  the  arcs  AB,  BC,  CD,  DA,  and  proceed 
as  before. 

By  repeating  this  process,  regular  inscribed  and  circum- 
scribed polygons  of  16,  32,..-,  and,  in  general,  of  2n  sides, 
may  be  drawn. 


130  PLANE   GEOMETRY.  [Bit.  V. 


EXERCISES. 

1.  Show  that  the  side  of  an  inscribed  square  =  r  v  2. 

2.  Find  the  ratio  of  the  areas  of  an  inscribed  and  a  circumscribed 
circle. 

PROPOSITION  V.     PROBLEM. 
296.    To  inscribe  in  a  given  circle  a  regular  hexagon. 


Let  0  be  the  centre  of  the  given  circle. 

To  inscribe  therein  a  regular  hexagon. 

Each  central  angle  of  an  inscribed  hexagon  will  be  one-sixth 
of  four  right  angles,  or  one-third  of  two  right  angles,  leaving 
(from  79)  two-thirds  of  two  right  angles  for  the  two  base  angles 
of  a  triangle  if  we  imagine  lines  drawn  from  the  centre  to  each 
vertex. 

But  these  lines  being  radii  are  equal,  forming  an  isosceles 
triangle. 

Therefore  the  angles  at  the  base  are  equal  or  each  will  be 
one-third  of  two  right  angles,  hence  all  the  angles  of  the  tri- 
angle are  equal  and  the  triangle  is  equilateral  or  the  base  is 
equal  to  the  radius. 

Therefore  apply  the  radius  six  times  to  the  circumference, 
join  the  points  of  division,  and  the  inscribed  figure  is  the 
hexagon  required. 

297.  COR.  1.  By  joining  the  alternate  vertices,  A,  C,  D,  an 
equilateral  triangle  is  inscribed  in  a  circle. 


§300.]  REGULAR  POLYGONS  AND   CIRCLES.  131 

298.  COR.  2.     By  bisecting  the  arcs  AB,  EC,  etc.,  a  regu- 
lar polygon  of  12  sides  may  be  inscribed  in  a  circle ;  and,  by 
continuing  the  process,  regular  polygons  of  24,  48,  etc.,  sides 
may  be  inscribed. 

PROPOSITION  VI.     THEOREM. 

299.  If  the  number  of  sides  of  a  regular  inscribed  polygon  be 
increased  indefinitely,  the  apothem  will  be  an  increasing  variable 
whose  limit  is  the  radius  of  the  circle. 


In  the  right  triangle  OCA,  let  OA  be  denoted  by  R,  OC  by 
r,  and  AC  by  6. 

To  prove  that  R  is  the  limit  of  r. 

In  the  triangle  OAC,  since  one  side  of  a  triangle  is  (by  78) 
greater  than  the  difference  between  the  other  two,  we  have 

R-r<b. 

Now  by  increasing  the  number  of  sides  each  side  diminishes 
in  length,  and  hence  the  half  side,  b,  can  by  increasing  the 
number  of  sides  indefinitely  be  made  less  than  any  assignable 
quantity,  or  the  difference  between  R  and  r  can  be  made  less 
than  any  assignable  quant  it}7",  hence  R  is  the  limit  of  r.  Q.E.D. 

PROPOSITION  VII.     THEOREM. 

300.  If  the  number  of  sides  of  a  regular  inscribed,  and  of 
a  similar  regular  circumscribed,  polygon,  is  indefinitely  increased, 

I.  Tlie  perimeter  of  each  polygon  approaches  the  circumfer- 
ence of  the  circle  as  a  limit. 


132  PLANE   GEOMETRY.  [BK.  V. 

II.    Their  areas  approach  the  area  of  the  circle  as  a  limit. 


' 


A  \      !£     / 

\     i     ' 

\   !   / 

\     I     / 

^V 


o 

1.  Let  AB  be  one  side  of  a  regular  inscribed  polygon,  ^.'ZJ' 
a  corresponding  side  of  a  regular  circumscribed  polygon  of  the 
same  number  of  sides,  and  0  the  centre  of  the  circle.  Call 
the  perimeter  of  the  inscribed  polygon  P,  and  of  the  circum- 
scribed, P'. 

To  prove  that  the  limit  of  P  and  P!  is  the  circumference  of 
the  circle. 

It  is  evident  that  the  inscribed  polygon  can  never  pass  with- 
out the  circle,  nor  can  the  circumscribed  polygon  come  within  ; 
therefore,  however  near  they  may  approach  one  another,  they 
will  be  still  nearer  the  circle. 

Since  the  polygons  are  regular,  they  are  (by  290)  similar, 


and  (by  291)  we  have  —  =        ,  but  (by  299)  the  difference  be- 

_P       OF 

tween  OE  and  OF,  when  the  number  of  sides  is  indefinitely 
increased,  approaches  0,  therefore  the  difference  between  P 
and  P'  will  approach  0. 

That  is,  the  perimeters  of  the  inscribed  and  circumscribed 
polygons  approach  one  another  ;  hence  each  will  approach  the 
circle  more  nearly. 

2.  Let  S  and  S'  represent  the  areas  of  the  inscribed  and  cir- 
cumscribed polygons. 

S'=OF*fOF\* 

S      OE2     \OEj' 

OF 

But  since  (by  299)  OF  approaches  OE,  -—  approaches  1,  or 

*  OE 

approaches  1. 


\_OE 


§  303.  ] 


REGULAR   POLYGONS   AND    CIRCLES. 


133 


Hence  -  -  approaches  1,  or  S'  approaches  S. 

S 
But  the  area  of  the  circle  being  interme'diate,  each  polygon 

will  approach  the  area  of  the  circle  more  nearly.  Q.E.D 

301.  DEFINITION.     In  circles  of  different  radii,  Similar  Arcs 
Segments,  or  Sectors  are  those  which  correspond  to  equal  central 
angles. 

PROPOSITION  VIII.     THEOREM. 

302.  The  circumferences  of  circles  have  the  same  ratio  as  their 
radii. 


Let  C  and  C'  be  the  circumferences,  R  and  R'  the  radii  of 
the  two  circles  0  and  0'. 

•To  prove  C :  C' :  :  R  :  R'. 

Inscribe  in  the  circle  two  regular  polygons  of  the  same  num- 
ber of  sides,  whose  perimeters  we  shall  call  P  and  P'. 

Then  by  (291) 

P      R 

P'     R'' 

Or  P  x  R'  =  P'  X  R. 

Suppose  the  number  of  sides  to  be  indefinitely  increased,  the 
two  to  continually  have  the  same  number  of  sides. 

Then  P  x  R'  will  approach  the  limit  C  x  R',  and  P'  x  R  will 
approach  the  limit  C"  x  R. 

Therefore  (by  171)  C  x  R'  =  C'  x  R,  or  -  Q.E.D. 


303.   COR.     By  multiplying  the  last  member  by  2  we  have 

C  =  2R 
C'~~2R') 


or  C:  C'  =  2R:2R'. 


134  PLANE   GEOMETRY.  [BK.  V. 

Taking  this  by  alternation  (202),  it  becomes 


2R     2R' 

That  is,  the  ratio  of  the  circumference  of  a  circle  to  its 
diameter  is  a  constant. 

This  constant  is  denoted  by  the  Greek  letter  IT, 

or  JL  =  7r?0r  C=27rR. 

2  .R 

EXERCISES. 

1.  Show  that  the  side  of  an  inscribed  equilateral  triangle  =  r  \/3. 

/\*         . — 

2.  Show  that  the  apotheni  of  a  regular  inscribed  hexagon  —  -  V 3. 

2 

3.  Find  the  area  of  a  square  inscribed  in  a  circle  whose  radius  is  6. 

4.  Show  that  the  area  of  a  regular  inscribed  hexagon  is  a  mean  propor- 
tional between  the  areas  of  an  inscribed,  and  of  a  circumscribed,  equi- 
lateral triangle. 

PROPOSITION  IX.     THEOREM. 

304.    The  area  of  a  circle  is  equal  to  one-half  the  product  of  its 
circumference  and  radius.     (Compare  292.) 


FOE 

Let  R  denote  the  radius,  C  the  circumference,  and  S  the 
area  of  the  circle. 

To  prove  that  S  =  ±C  x  R. 

Circumscribe  about  the  circle  a  regular  polygon ;  let  P  de- 
note its  perimeter  and  P1  its  area. 

Then  (by  292),  P'  =  ±Px 


§307.]  REGULAR   POLYGONS  AND   CHICLES.  135 

But  when  the  number  of  sides  of  the  polygon  increases 
indefinitely,  P  approaches  C  (by  300),  OG  remains  It,  and  P' 
approaches  S. 

Therefore  £  =  |  R  x  C.  Q.E.D. 

305.   COR.  1.  C=27rR  (by  303). 

Therefore  S  =     Rx2icR 


306.  COR.  2.     Since  a  sector  bears  the  same  ratio  to  the  circle 
that  its  arc  bears  to  the  circumference,  the  area  of  a  sector  is 
equal  to  one-half  the  product  of  its  arc  by  its  radius. 

PROPOSITION  X.     PROBLEM. 

307.  Given  the  raditts  and  the  side  of  a  ret/tilar  inscribed  poly- 
gon, to  compute  the  side  of  a  xi'mildr  circumscribed  polygon. 


Let  AB  be  a  side  of  the  inscribed  polygon,  and  OF '=  R,  the 
radius  of  the  circle. 

To  compute  CD,  a  side  of  the  similar  circumscribed  polygon. 

Draw  CO  and  DO;  they  will  (by  286)  intersect  AB  in  A 
and  B. 

The  triangles  CFO  and  AEO  are  corresponding  parts  of 
similar  polygons,  and  hence  are  similar. 

Hence  CF=OF 

AE     OE 


136 


PLANE   GEOMETRY. 


[BK.  V, 


or 


Multiplying  by  AE,      CF= 

CD  = 


OF  x  AE     R  x  AE 


OE 

RxAB 
OE 


OE 


In  the  right  triangle  OAE  (by  266) 


R2  -  AB'. 


Therefore 


CD  = 


2RxAB 


V4  R*  - 


PROPOSITION  XI.     PROBLEM. 

308.  Given  the  radius  and  the  side  of  a  regular  inscribed 
polygon,  to  compute  the  side  of  the  regular  inscribed  polygon  of 
double  the  number  of  sides. 

Given  AB,  a  side  of  the  regular  inscribed  c 

polygon,  and  OC  —  R,  the   radius  of   the 
circle. 

To  compute  AC,  a  side  of  an  inscribed 
polygon  of  double  the  number  of  sides. 

Draw  OC  bisecting  arc  ACB ;  then  since 
it  bisects  the  arc  ACB,  it  will  bisect  AB  at 
right  angles  (by  147). 

Produce  CO  to  D  and  draw  AD ;  then  since  Z.  CAD  is  in- 
scribed in  a  semicircle  (by  177),  it  is  a  right  angle. 

Then  (by  228)  AC  is  a  mean  proportional  between  CD  and 
CE,  or 

AC2  =  CDxCE  =  CD(CO  -  -  EO)=  2  R(R--  EG) 

=  R(2R-2EO). 
But  in  the  right  triangle  (by  266), 

EO  =  V  OA2  -  AE2  =\  R2  -  —  =  i  V4  R2  -  AB\ 
or         2EO  =  V4  R2  -  AB2. 


§309.]  REGULAR  POLYGONS  AND   CIRCLES.  137 


Substituting  this  in  the  value  for  AC',  we  have 


or 


-  V4  R2  - 


PROPOSITION  XII.     PROBLEM. 

309.  To  compute  the  ratio  of  the  circumference  of  a  circle  to 
its  diameter. 

It  is  known  (from  303)  that 

C  =  2  TrR. 

If,  therefore,  we  take  a  circle  whose  radius  is  unity,  we  have 

<7  =  27r,    or   7r  =  i<7; 

that  is,  TT  =  a  semicircumference  of  unit  radius. 

Hence  the  semiperimeter  of  each  inscribed  polygon  is  an 
approximate  value  of  TT,  and  the  semiperimeter  of  each  circum- 
scribed polygon  is  also  an  approximate  value  of  TT.  Therefore, 
if  by  constantly  increasing  the  number  of  sides  of  these  poly- 
gons the  approximate  values  of  TT  become  practically  identical, 
we  know  that  as  the  circle  lies  between  the  inscribed  and  cir- 
cumscribed polygons,  this  coincident  value  for  ?r  can  be  taken 
as  the  semicircumference  of  the  circle  of  unit  radius. 

If  we  begin  with  the  square  we  know  that  the  side  is  the 
hypotenuse  of  an  isosceles  right  triangle,  the  two  equal  sides 
being  radii  ;  hence 

AB  (in  308)  =  V2  =  1.4142136, 
or  semiperimeter  =  2.8284272. 

Then  each  side  of  the  circumscribed  square  is  the  diameter 
or  twice  the  radius  =  2,  and  the  semiperimeter  Avill  be  4. 

From  the  final  equation  in  Prob.  XI.  it  is  easy  to  compute 
the  semiperimeter  of  a  polygon  of  8  sides  ;  then  from  the  final 
equation  in  Prob.  X.  can  be  computed  the  semiperimeter  of  a 
circumscribed  polygon  of  8  sides  ;  and  so  on. 


138 


PLANE   GEOMETRY. 


[Bit.  V. 


In  the  following  table  are  given  the  semiperimeters  of  in- 
scribed and  circumscribed  polygons : 


NUMBER  OF  SIDES. 

INSCRIBED. 

CIRCUMSCRIBED. 

4 

2.8284271 

4.0000000 

8 

3.0616675 

3.3137085 

16 

3.1214452 

3.1825927 

32 

3.1365485 

3.1517249 

64 

3.1403312 

3.1414184 

128 

3.1412773 

3.1422236 

256 

3.1415138 

3.1417504 

512 

3.1415729 

3.1416321 

1024 

3.1415877 

3.1416025 

2048 

3.1415914 

3.1415951 

4096 

3.1415923 

3.1415933 

8192 

3.1415926 

3.1415928 

The  figures  in  face  type  show  the  approximation. 

310.  SCHOLIUM.  By  the  aid  of  simpler  methods  the  value 
of  TT  has  been  computed  to  more  than  eight  hundred  places  of 
decimals. 

The  first  twenty  figures  of  the  result  are 

TT  =  3.14159  26535  89793  238, 

-  =  0.31830  98861  83790  6715, 
log  TT  =  0.49714  98726  94133  85435. 
For  all  practical  purposes  it  is  sufficient  to  take  TT  =  3.1416. 

EXERCISES. 

1.  If  the  radius  of  a  circle  is  4,  find  its  circumference  and  area. 

2.  If  the  circumference  of  a  circle  is  30,  find  its  radius  and  area. 

3.  If  the  diameter  of  a  circle  is  26,  find  the  length  of  an  arc  of  72°. 

4.  If  the  radius  of  a  circle  is  12,  find  the  area  of  a  sector  whose  central 
angle  is  80°. 

5.  If  the  apothem  of  a  regular  hexagon  is  4,  find  the  area  of  the 
circumscribing  circle. 


§  313.] 


REGULAR   POLYGONS  AND   CIRCLES. 


139 


MAXIMA  AND  MINIMA. 

311.  Of  quantities  of  the  same  kind,  the  one  which  is  the 
greatest  is  called  the  Maximum,  and  the  least  is  called  the 
Minimum. 

312.  Isoperimetric  figures  are  those  which  have  equal  perim- 
eters. 

PROPOSITION  XIII.     THEOREM. 

313.  Of  all  triangles  formed  icith  two  given  sides,   that   in 
which  these  sides  include  a  right  angle  is  the  maximum. 


B 


D 


Let  ABC  and  ABC  be  two  triangles  having  the  sides  AB 
and  BC  equal  to  the  sides  AB  and  BC  respectively,  and  let 
the  angle  ABC  be  a  right  angle. 

To  prove  that 

area  ABC  >  area  ABC. 

Draw  AD  perpendicular  to  BC. 

Then  since  (by  52)  the  oblique  line  AB  is  greater  than  the 
perpendicular  AD,  we  have 

AB  >  AD. 

But  AB  and  AD  are  the  altitudes  of  the  triangles  ABC  and 
ABC,  and  as  they  have  the  same  base,  that  triangle  is  the 
greater  which  has  the  greater  altitude,  or 


area  ABC  >  area  ABC. 


Q.  E.  D. 


140 


PLANE  GEOMETRY. 


[BK.  V. 


PROPOSITION  XIV.     THEOREM. 

314.    Of  isoperimetric  triangles  having   the  same  base,  that 
which  is  isosceles  is  the  maximum. 


Let  ABC  and  A'BC  be  two  isoperimetric  triangles  having 
the  same  base  BC,  and  let  the  triangle  ABC  be  isosceles. 

To  prove  that         area  ABC  >  area  A'BC. 

Produce  AB  to  D,  making  AD  =  AB,  and  draw  CD. 

Since  B,  C,  and  D  are  equally  distant  from  A,  a  circle  with 
A  as  a  centre  could  be  drawn  through  B,  C,  and  D,  of  which 
BD  would  be  the  diameter. 

The  angle  BCD  would  therefore  (by  177)  be  a  right  angle. 

Draw  AF  and  A'G  parallel  to  BC,  take  A'E  equal  to  A'C, 
and  draw  BE. 

Since  the  triangles  ABC  and  A'BC  are  isoperimetric, 

AB  +  AC  =  A'B  +  A'C  =  A'B  +  A'E. 
But  AC=AB  = 

hence  A'B  +  ^'^  =  BD. 

But  (by  6)  A'B  +  ^'#  >  £# ; 

that  is,  BD  >  BE. 

Therefore  (by  58)  CD  >  CE. 


§310.]  REGULAR   POLYGONS  AND   CIRCLES.  141 

Since  the  triangles  CAD  and  CA'E  are  isosceles  by  con- 
struction, and  AF  and  AG  perpendiculars  upon  their  bases, 


CF=±CD,  and 


= 


But  as  CD  is  greater  than  CE,  CF>  CG. 

CF  is  the  altitude  of  the  triangle  BAG,  and  CG  of  BAG, 
as  these  triangles  have  the  same  base,  the  one  which  has  the 
greater  altitude  is  the  greater,  or 

area  ABC  >  ABC.  Q.E.D. 

315.  COR.  Of  all  the  triangles  of  the  same  perimeter,  that 
which  is  equilateral  is  the  maximum. 

For  the  maximum  triangle  having  a  given  perimeter  must 
be  isosceles  whichever  side  is  taken  as  the  base. 


PROPOSITION  XV.     THEOREM. 

316.    Of  isoperimetric  polygons  having  the  same  number  of 
sides,  that  tvhich  is  equilateral  is  the  maximum. 


B    B' 


Let  ABCDE  be  an  equilateral  polygon. 
To  prove  that  it  is  greater  than   any  other  isoperimetric 
polygon  of  the  same  number  of  sides. 

If  not  greater,  suppose  AB'CDE  is  greater. 

Draw  AC. 

Then  ABC  being  an  isosceles  triangle,  we  know  (by  314) 


142  PLANE   GEOMETRY.  [BK.  V. 

Add  area  ACDE  to  this  inequality, 

ACDE  +  ABC  >  ACDE  +  AB'C, 
or  ABODE  >  AB'CDE. 


That  is,  AB  and  5Gy  cannot  be  unequal  without  decreasing  the 
area,  and  in  like  manner  it  can  be  shown  that  BC  =  CD  =  DE, 
etc.,  or  the  polygon  is  equilateral.  Q.E.D. 

PROPOSITION  XVI.     THEOREM. 

317.  Of  two  isoperimetric  regular  polygons,  that  which  has  the 
greater  number  of  sides  has  the  greater  area. 


N 


A  DC 

Let  M  be  an  equilateral  triangle,  and  N  an  isoperimetric 
square. 

To  prove  that  area  N  >  area  M. 

Let  I)  be  any  point  in  the  side  AC  of  the  triangle. 

Then  the  triangle  M  may  be  regarded  as  an  irregular  quad- 
rilateral, having  the  four  sides  AB,  BC,  CD,  and  DA\  the 
angle  at  D  being  equal  to  two  right  angles. 

Hence,  since  the  two  quadrilaterals  are  isoperimetric, 

area  N  >  area  M.  (316) 

In  like  manner,  it  may  be  proved  that  the  area  of  a  regular 
pentagon  is  greater  than  that  of  an  isoperimetric  square;  that 
the  area  of  a  regular  hexagon  is  greater  than  that  of  an  isoperi- 
metric regular  pentagon ;  and  so  on.  Q.E.D. 


§  310.] 


REGULAR   POLYGONS  AND   CIRCLES. 


143 


318.  COR.  Since  a  circle  may  be  regarded  as  a  regular  poly- 
gon of  an  infinite  number  of  sides,  it  follows  that  the  circle  is  the 
maximum  of  all  isoperimetric  plane  figures. 


EXERCISES. 

1.    Of  all  triangles  of  given  base  and  area, 
the  isosceles  is  that  which  has  the  greatest 


vertical  angle. 


2.  The  shortest  chord  which  can  be  drawn 
through  a  given  point  within  a  circle  is  the 
perpendicular  to  the  diameter  which  passes 
through  that  point. 


PROPOSITION  XVII.     THEOREM. 

319.  The  sum  of  the  distances  from  two  fixed  points  on  the 
same  side  of  a  straight  line  to  the  same  point  in  that  line  is  a 
minimum  when  the  lines  joining  the  fixed  points  with  the  same 
point  are  equally  inclined  to  the  given  line. 


C    F. 


L   D 


Q 


Let  CD  be  the  straight  line,  A  and  B  the  fixed  points,  P 
such  a  point  in  CD  that  Z  APC  =  Z.  BPD,  and  Q  any  other 
point  in  CD. 


144 


PLANE   GEOMETRY. 


[BK.  V.,  §  319. 


To  prove  that         AP+PB <  AQ,  +  BQ. 

Let  fall  the  perpendicular  AF,  and  continue  it  until  it  meet 
BP  produced,  say  in  E,  and  join  QA  and  QE. 
Since  BE  and  CD  are  intersecting  lines  (by  49), 

Z  BPD  =  Z  FPE. 

By  hypothesis,          Z  .1PP  =  Z  £PZ> ; 
therefore  Z  ^IPP  =  Z  FPE. 

The  triangles  ^IPP  and  PP#  are  right  triangles  by  con- 
struction, and  having  the  side  FP  common,  are  (by  90)  equal 
in  all  their  parts ;  that  is 

AF  =  FE,  and  AP  =  EP. 

Then  (by  53)  AQ  =  EQ. 

But  (by  6)  EB<EQ+  QB} 


or 


Therefore 


QB. 


Q.E.D. 


NOTE.  If  CD  is  a  reflecting  surface,  a  ray  of  light  in  order  to  go  from 
A  to  B  by  reflection,  pursues  the  shortest  path  when  the  angle  of  inci- 
dence (APF}  is  equal  to  the  angle  of  reflection  (BPD}.  This  is  the 
physical  law,  thus  furnishing  one  illustration  of  the  economy  in  nature. 


EXERCISES 


1.  Given  the  base  and  the  verti- 
cal angle  of  a  triangle ;  to  construct 
it  so  that  its  area  may  be  a  maxi- 
mum. 

SUGGESTIOX.     See  195,  Ex.  1. 

2.  Show  that    the    greatest    rec- 
tangle which  can  be  inscribed  in  a 
circle  is  a  square. 


SOLID    GEOMETRY. 

BOOK   VI. 

PLANES  AND  SOLID  ANGLES. 
DEFINITIONS. 

320.  A  plane  is  (by  9)  a  surface  such  that  a  straight  line 
which  joins  any  two  of  its  points  will  lie  wholly  in  the  surface. 

321.  A  plane  is  of  unlimited  extent  in  its  length  and  breadth ; 
but  to  represent  a  plane  in  a  diagram  it  is  necessary  to  take 
only  a  definite  portion,  and  usually  it  is  represented  by  a  par- 
allelogram which  is  supposed  to  lie  in  the  plane. 

322.  A  plane  is  said  to  be  determined  by  any  combination  of 
lines  or  points  when  it  is  the  only  plane  which  contains  these 
lines  or  points. 

323.  Any  number  of  planes  may  be  passed  through  any  given 
straight  line. 

For  if  a  plane  is  passed   through  any 
given  straight  line  AB,  the  plane  may 
be  turned  about  AB  as  an  axis,  and  made      , 
to  occupy  an  infinite  number  of 'positions,      /> 
each  of  which  will  be  a  different  plane    *+ 
passing  through  AB. 

From  this  it  can  be  seen  that  a  single  straight  line  does  not 
determine  a  plane. 

145 


146  KOLII)   GEOMETRY.  [BK.  VI. 

_ 

324.  But  a  plane  is  determined  by  a  straight  line  and  a  point 

without  that  line.  

r 

For,  if  tho  plane  containing  the  straight 
line  AB  turn  about  this  line  as  an  axis  until      A- 
it  contains  tin*.  given  point  (7,  the  plane  is 
evidently  determined,  for  if  turned  in  any  other  position  it 
will  not  contain  C. 

325.  A  plane  is  determined  by  a  straight  line  and  a  point 
without  that  line,  by  two  intersecting  straight  lines,  or  by  two 
parallel  lines. 

Since  one  straight  line  and  a  point  without  that  line  deter- 
mine (by  324)  the  plane,  it  will  be  necessary  to  take  only  three 
points,  two  in  one  of  the  lines  and  the  third  in  the  other  line, 
the  first  two  giving  the  required  line  and  the  third  the  required 
point, 

326.  A  straight  line  is  perpendicular  to  a  plane  when  it  is 
perpendicular  to  every  straight  line  of  the  plane  which  passes 
through  its/oo^,  that  is,  the  point  where  it  meets  the  plane. 

Conversely,  the  plane  is  perpendicular  to  the  line. 

327.  A  straight  line  is  said  to  be  parallel  to  a  plane  when 
they  cannot  meet,  however  far  they  may  be  produced. 

328.  Two  planes  are  said  to  be  parallel  to  each  other  when 
they  cannot  meet,  however  far  they  may  be  produced. 

329.  The  projection  of  a,  point  on  a  plane  is  the  foot  of  the 
perpendicular  let  fall  from  the  point  to  the  plane. 

330.  The  projection  of  a  line  on  a  plane  is  the  line  through 
the  projections  of  all  its  points. 

331.  The  angle  which  a  line  makes  with  a  plane  is  the  angle 
which  it  makes  with  its  projection  on  the  plane. 

332.  By  the  distance  of  a  point  from  a  plane  is  meant  the 
shortest  distance  from  the  point  to  the  plane. 


§  334.  ] 


PLANES  AND   SOLID  ANGLES. 


147 


PROPOSITION  I.     THEOREM. 

333.    If  two  planes  cut  each  other,  their  com  mo  it  intersection  is 
a  straight  line. 

Let  AB,  CD  be  two  planes  which 
cut  each  other. 

To  prove  their  common  intersec- 
tion is  a  straight  line. 

Let  H  and  E  be  two  points  in 
the  intersection.  Join  them  by  the 
straight  line  HE. 

By    definition   this   straight  line 
lies  wholly  in  the  plane  AB,  like- 
wise H  and  E  being  points  on  CD  the  line  HE  must  lie  wholly 
in  the  plane  CD. 

Therefore   HE  being  common  to  both    planes,  it  must  be 
their  intersection.  Q.E.D. 


PROPOSITION  II.     THEOREM. 

334.    If  oblique  lines  are  drawn  from  a  point  to  a  plane: 

(1)  Two  oblique  lines  meeting  the  plane  at  equal  distances 
from  the  foot  of  the  perpendicular  are  equal. 

(2)  Of  two  oblique  lines  meeting  the  plane  at  unequal  distances 
from  the  foot  of  the  perpendicular,  the  more  remote  is  the  longer. 

Let  OP  be  perpendicular  to  the 
plane  MN,  and  PA  =  PB,  but 
PC  >  PA. 

To  prove  that  OA  =  OB,  but  that 
OC  >  OA. 

In  the  two  right  triangles   OP  A 
and  OPB,  the  side  OP  is  common 
and  AP  =  PB  by  hypothesis ;  there- 
fore (by  86)  the  triangles  are  equal    Mi 
in  all  their  parts,  that  is  OA  =  OB. 


148 


SOLID    GEOMETRY. 


[Bit.  VI. 


Since  OC  meets  the  line  PB  pro- 
duced at  a  point  further  from  the 
point  P  than  does  OB,  OC  is  (by  58) 
greater  that  OB. 

But  OB  =  OA, 

therefore        OC  >  OA.  Q.E.D. 

335.  COR.  1.    The  perpendicular  is 
£/<e   shortest   distance  from  a  point  to 

a  plane;  therefore,  by  the  distance  of  a  point  from  a  plane  is 
meant  the  perpendicular  distance  from  the  point  to  the  plane. 

Since  OP  is  less  than  OA,  OB,  and  OC,  it  is  the  shortest 
distance  from  the  point  to  the  plane. 

336.  COR.  2.    Equal  oblique  lines  from  a  point  to  a  plane  meet 
the  plane  at  equal  distances  from  the  foot  of  the  perpendicular; 
and  of  two  unequal  oblique  lines,  the  greater  meets  the  plane  at 
the  greater  distance  from  the  foot  of  the  perpendicular. 

PROPOSITION   III.     THEOREM. 

337.  If  a  straight  line  is  perpendicular  to  each  of  two  straight 
lines  at  their  point  of  intersection  it  is  perpendicular  to  the  plane 
of  those  lines. 


Let  BA  be  perpendicular  to  AD  and  AC  two  intersecting 
lines  in  the  plane  MN,  and  let  EA  be  any  other  line  in  JfJV 
passing  through  the  point  of  intersection  of  AD  and  AC. 


§343.]  PLANES  AND   SOLID  ANGLES.  149 

To  prove  that  BA  is  perpendicular  to  EA  and  hence  per- 
pendicular to  MN. 

Make  AC  equal  to  AD,  draw  DC,  and  produce  BA  to  B', 
making  AB'  =  AB,  and  join  B  with  D,  E,  and  C. 

Since  D  and  C  are  equally  distant  from  A,  BD  —  BC  (by 
334). 

Since  ZL4  is  a  perpendicular  bisector  of  BB',  (by  54)  DB'  = 
BD,  likewise  B'C '=  BC,  hence  B'DC  is  an  isosceles  triangle. 

The  triangles  BDC  and  BDC  have  BD  =  B'D,  BC=B'C, 
and  the  side  DC  common;  they  are  therefore  equal  (by  9.1)  in 
all  their  parts. 

Therefore  if  the  triangle  B'DC  were  applied  to  BDC  they 
would  coincide  in  all  their  parts,  and  the  point  E  being  fixed, 
the  line  BE  would  fall  upon  B'E  and  be  equal  to  it. 

Hence  E  being  equally  distant  from  B  and  B',  it  is  on  the 
perpendicular  bisector  of  BB',  or  EA  is  perpendicular  to  BA, 
or  BA  is  perpendicular  to  AE. 

As  A  E  is  any  line  in  MN,  BA  is  perpendicular  to  MN.  Q.E.D. 

338.  COR.  1.    Conversely,  all  the  perpendiculars  to  a  straight 
line  at  the  same  point  lie  in  a  plane  perpendicular  to  the  line. 

339.  COR.  2.   At  a  given  point  in  a  plane,  only  one  perpen- 
dicular to  the  plane  can  be  erected. 

340.  COR.  3.   From  a  point  without  a  plane  only  one  perpen- 
dicular can  be  drawn  to  the  plane. 

341.  COR.  4.   At  a  given  point  in  a  straight  line  one  plane, 
and  only  one,  can  be  drawn  perpendicular  to  the  line. 

342.  COR.  5.    If  a  right  angle  be  turned  round  one  of  its  arms 
as  an  axis,  the  other  arm  idll  generate  a  plane. 

343.  COR.  6.    Through  a  given  point  without  a  straight  line 
one  plane,  and  only  one,  can  be  drawn  perpendicular  to  the  line. 


150 


SOLID   GEOMETRY. 


[BK.  VI. 


PROPOSITION  IV.     THEOREM. 

344.  If  through  the  foot  of  a  perpendicular  to  a  plane  a  line 
is  drawn  at  rigid  angles  to  any  line  in  the  plane,  the  line  drawn 
from  its  intersection  with  this  line  to  any  point  in  the  perpen- 
dicular will  be  perpendicular  to  the  line  in  the  plane. 

~ 


Let  AB  be  a  perpendicular  to  the  plane  MN. 

Draw  AE  perpendicular  to  any  line  CD  in  the  plane  MN, 
and  join  the  point  E  to  any  point  B  in  the  perpendicular. 

To  prove  that  BE  is  perpendicular  to  CD. 

Take  EC  =  ED,  and  draw  AD,  AC,  BD,  and  BC. 

Since  A  is  on  the  perpendicular  bisector  of  CD,  AD  =  AC 
(by  54). 

Hence  the  oblique  lines  BD  and  BC  meet  the  plane  MN  at 
points  equally  distant  from  the  foot  of  the  perpendicular  and 
are  (by  334)  equal. 

Therefore  BDC  is  an  isosceles  triangle,  and  the  line  BE 
bisecting  the  base,  by  construction,  will  be  (by  94)  perpendicu- 
lar to  the  base ;  that  is,  Z  BEG  is  a  right  angle.  Q.E.D. 

EXERCISES. 

1.  If  a  plane  bisects  a  straight  line  at  right  angles,  every  point  in  the 
plane  is  equally  distant  from  the  extremities  of  the  line. 

2.  Given  a  plane  MN  and  two  points  A  and  7>  on  the  same  side  of  the 
plane,  find  upon  the  plane  a  point  C  so  that  the  sum  of  the  distances  AC 
and  BC  shall  be  a  minimum. 

3.  If  the  points  are  on  opposite  sides,  find  C  when  the  difference  of  the 
distances  is  a  minimum. 


§347.]  PLANES  AND   SOLID  ANGLES.  151 

PROPOSITION  V.     THEOREM. 

345.  Two  straight  lines  perpendicular  to  the  same  plane  are 
parallel. 

Let  AB  and  CD  be  two  straight  lines 
perpendicular  to  the  plane  MN. 

To  prove  that  AB  and  CD  are  paral- 
lei. 

In  the  plane  MN  draw  BD  and  AD 
and  erect  DE  perpendicular  to  BD. 

Since   DC   is    perpendicular,    to    the 
plane  MN  it  is  (by  326)  perpendicular  to  DE. 

Again,  since  BD  is  perpendicular  to  ED,  by  construction, 
AD  is  perpendicular  (by  344)  to  ED. 

Therefore  ED  is  perpendicular  to  AD,  BD,  and  CD ;  hence 
these  lines  all  lie  in  one  plane. 

Consequently  AB  and  CD  are  two  lines  in  one  plane  perpen- 
dicular to  the  same  line  BD,  therefore  (by  61)  they  are  parallel 
to  one  another.  Q.E.D. 

346.  COR.  1.     If  one  of  ttvo  parallels  is  perpendicular  to  a 
plane,  the  other  is  also. 

347.  COR.  2.     Two  straight  lines  that  are  parallel  to  a  third 
straight  line  are  parallel  to  each  other. 

EXERCISES. 

1.  Two  planes  that  have  three  points  not  in  the  same  straight  line 
in  common  coincide. 

2.  At  a  given  point  in  a  plane,  erect  a  perpendicular  to  the  plane. 

3.  From   a  point  without   a  plane,  let  fall   a  perpendicular  to  the 
plane. 

4.  From  a  point  without  a  plane  draw  a  number  of  equal  oblique  lines 
to  the  plane. 


152  SOLID   GEOMETRY.  [Bs.  VI. 


PROPOSITION  VI.     THEOREM. 

348.   If  a  straight  line  and  a  plane  be  perpendicular  to  the 
same  straight  line,  they  are  parallel. 

B  C 


D       / 
Jff 


Let  the  straight  line  BC  and  the  plane  MN  be  perpen- 
dicular to  the  straight  line  AB. 

To  prove  that  BC  is  parallel  to  MN.  Pass  a  plane  through 
BC  and  A  meeting  MN  in  the  line  AD,  then  from  any  point 
C  in  the  line  BC  let  fall  the  perpendicular  CD  to  the  plane, 
and  draw  AD. 

Since  CD  is  perpendicular  to  MN  it  will  (by  326)  be  per- 
pendicular to  AD. 

But  BA  is  perpendicular  to  AD,  therefore  (by  345)  BA  is 
parallel  to  CD,  and  likewise  BC  and  AD  being  perpendicular 
to  BA,  they  will  be  parallel. 

Therefore  BADC  is  a  parallelogram  and  CD  =  BA,  or  the 
line  BC  is  everywhere  equally  distant  from  MN,  hence  is 
parallel  to  MN.  Q.E.D. 

349.  COR.  1.     If  two  planes  be  perpendicular  to  the  same 
straight  line,  they  are  parallel. 

350.  COR.  2.      Two  parallel  planes   are   everywhere   equally 
distant.     And,  conversely,  two  planes  that  are  everywhere  equally 
distant  are  parallel. 

351.  COR.  3.     If  tivo  intersecting  straight  lines  are  each  par- 
allel to  a  given  plane,  the  plane,  of  these  lines  is  parallel  to  the 
given  plane. 


§  353.] 


PLANES  AND   SOLID  ANGLES. 


153 


PROPOSITION  VII.     THEOREM. 

352.  The  intersections  of  two  parallel  planes  by  a  third  plane 
are  parallel  lines. 

Let  MN  and  PQ  be  two  parallel  planes 
intersected  by  the  plane  AD  in  AB  and 
CD. 

To  prove  that  AB  and  CD  are  parallel. 

The  lines  AB  and  CD  cannot  meet  since 
they  lie  in  planes  that  are  parallel,  they 
themselves  by  hypothesis  being  in  the 
same  plane  AD. 

Therefore  AB  and  CD  are  parallel. 

EXERCISES. 

1.  If  a  straight  line  is  parallel  to  a  line  in  a  plane,  it  is  parallel  to  the 
plane. 

2.  Parallel  lines  between  parallel  planes  are  equal. 
SUGGESTION.     See  108. 

PROPOSITION  VIII.     THEOREM. 

353.  If  two  angles  not  in  the  same  plane  have  their  sides 
respectively  parallel  and  lying  in   the  same  direction,  they  are 
equal  and  their  planes  are  parallel. 

Let  the  angles  C  and  C"  lie  in  the  planes 
MN  and  PQ,  respectively,  having  their 
sides  AC  and  A'C'  parallel,  and  also  CB 
and  C'B  parallel  and  in  the  same  direction. 

To  prove  that  Z  C  =  Z  C",  and  that  MN 
and  PQ  are  parallel. 

1.  Take  A'C'  =  AC  and  C'B'  =  CB,  and 
draw  AA,  BB',  and  CC'. 

Since  AC  and  A'C'  are  equal  by  con- 
struction and  parallel  by  hypothesis,  the  figure  AGO  A  is  a 
parallelogram ;   that  is,  A  A1  is  equal  and  parallel  to  CO. 


154 


SOLID   GEOMETRY. 


[BK.  VI. 


For  a  similar  reason,  BB'  and  CC'  are  equal  and  parallel. 

Since  AA'  and  BB'  are  both  equal  and  parallel  to  (7(7',  they 
are  equal  and  parallel  to  each  other,  or  ABB1  A'  is  a  parallelo- 
gram, and  hence  AB  =  A'B'. 

Therefore  the  triangles  ACB  and  A'C'B'  have  their  sides 
respectively  equal,  and  hence  their  angles  are  equal,  or 
Z  C  =  Z  C". 

2.  Since  AA  =  BB'  =  CO',  the  two  planes  are  equally  dis- 
tant, and  hence  parallel.  Q.E.D. 

354.  COR.     If  two  angles  have  their  sides  parallel,  they  are 
equal  or  supplemental. 

PROPOSITION  IX.     THEOREM. 

355.  If  two  straight  lines  be  intersected  by  three  parallel  planes 
their  corresponding  segments  are  proportional. 


Let  AB  and  CD  be  intersected  by  the  parallel  planes  MN, 
PQ,  ItS,  in  the  points  A,  E,  B,  and  C,  F,  D. 

AE      CF 


We  are  to  prove 


EB  " '  FD 


Draw  AD,  cutting  the  plane  PQ  in  G. 
Join  the  points  E,  G,  and  F,  G. 


§359.]  DIEDRAL   ANGLES.  15.5 

In  the  triangle  ABD,  EG,  being  in  the  plane  PQ  parallel  to 
jR>Sy,  will  be  parallel  to  BD. 

AE     AG 


Therefore  (by  213) 


EB"  GD 


Likewise  in  the  triangle  DAC,  GF  is  parallel  to  AC,  and 

hence 

CF  =  AG 

FD      GD 

Hence  (by  28) 


356.  COR.     Any  number  of  straight  lines  cut  by  parallel  planes 
are  divided  into  proportional  segments. 

DIEDRAL  ANGLES. 

DEFINITIONS. 

357.  When  two  planes  intersect  they  are  said  to  form  with 
each  other  a  Diedral  Angle. 

The  line  of  intersection  is  called  the  Edge.     The  planes  are 
the  Faces. 

Thus  in  the  diedral  angle  formed  by  the 
planes  BD  and  BF,  BE  is  the  edge  and  BD 
and  BF  are  the  faces. 

358.  A  diedral  angle  may  be  designated  by 
two  letters  on  its  edge ;  or,  if  several  diedral 
angles  have  a  common  edge,  by  four  letters, 
one  in  each  face  and  two  on  the  edge,  the  let- 
ters on  the  edge  being  named  between  the  other  two. 

Thus  the   diedral   angle    in   the  figure   may  be   designated 
either  as  BE  or  A-BE-C. 

359.  If  a  point  is  taken  in  the  edge  of  the  diedral  angle, 
and  two  straight  lines  are  drawn  through  this  point,  one  in 


156  SOLID   GEOMETRY.  [BK.  VI. 

each  face,  and  each  perpendicular  to  the  edge,  the  angle  formed 
by  these  two  lines  is  called  the  Plane  Angle  of  the  diedral 
angle,  as  Z  GHK. 

360.  Two  diedral  angles  are  equal  if  their  plane  angles  are 
equal,  or  when  their  faces  may  be  made  to  coincide. 

3S1.  The  magnitude  of  a  diedral  angle  depends  solely  on  the 
amount  of  divergence  of  its  faces,  and  is  entirely  independent 
of  their  extent. 

362.  Two  diedral  angles  are  adjacent  when  they  have   a 
common  edge  and  a  common  face  between  them. 

363.  When  the  adjacent  diedral  angles  which  a  plane  forms 
with  another  plane  on  opposite  sides  are  equal,  each  of  these 
angles  is  called  a  right  diedral  angle;  and  the  first  plane  is  said 
to  be  perpendicular  to  the  other. 

Thus    if    the    adjacent    diedral    angles  A 

ABCM,  ABCN  are  equal,  each  of  these  is 
a  right  diedral  angle,  and  the  planes  AC 

JiJt  J3 

and  MN  are  perpendicular  to  each  other. 

Through  a  given  line  in  a  plane  only 
one  plane  can  be  passed  perpendicular  to 
the  given  plane. 


H 


N 


364.  If  the  diedral  angle  is  a  right  angle,  the  plane  angle  is 
also  a  right  angle:  therefore  if  two  planes  are  perpendicular 
to  each  other,  a  straight  line  drawn  in  one  of  them  perpendicu- 
lar to  their  intersection  is  perpendicular  to  the  other. 

365.  The  following  principles  are  also  true : 

1.  If  a  straight  line  is  perpendicular  to  a  plane,  every  plane 
passed  through  the  line  is  perpendicular  to  that  plane. 

2.  If  two  planes  are  perpendicular  to  each  other,  a  straight 
line  through  any  point  of  their  intersection  perpendicular  to 
one  of  the  planes  will  lie  in  the  other. 


§368.]  DIEDRAL  ANGLES.  157 

3.  If  two  planes  are  perpendicular  to  each  other,  a  straight 
line  from  any  point  of  one  plane  perpendicular  to  the  other 
will  lie  in  the  first  plane. 

366.  Vertical  diedral  angles  are  those  which  have  a  common 
edge,  and  the  faces  of  one  are  prolongations  of  the  faces  of  the 
other. 

367.  Diedral  angles  are  acute,  obtuse,  complementary,  supple- 
mentary, under  the  same  conditions  that  hold  for  plane  angles. 

368.  The   demonstrations    of    many   properties    of   diedral 
angles   are    the    same    as    the    demonstrations   of    analogous 
properties  of  plane  angles. 

For  example : 

1.  Vertical  diedral  angles  are  equal. 

2.  Diedral   angles  whose  faces  are  respectively  parallel  or 
perpendicular  are  either  equal  or  supplementary. 

3.  Every  point  in  the  bisecting  plane  of  a  diedral  angle  is 
equally  distant  from  the  faces  of  the  angle. 

4.  If  a  plane  meets  another,  the  sum  of  the  adjacent  diedral 
angles  formed  is  equal  to  two  right  diedral  angles ;  and  con- 
versely. 

5.  If  two  parallel  planes  are  cut  by  a  third  plane,  the  alter- 
nate-interior diedral  angles   are    equal,  the    alternate-exterior 
angles  are  equal,  any  diedral  angle  is  equal  to  its  correspond- 
ing angle,  and  the  sum  of  the  interior  diedral  angles  on  the 
same  side  of  the  secant  plane  is  equal  to  two  right  diedral 
angles;   and  conversely. 

6.  Two  diedral  angles  whose  faces  are  parallel  each  to  each 
are  either  equal  or  together  equal  to  two  right  diedral  angles. 

7.  Diedral  angles  are  to  each  other  as  their  plane  angles ; 
hence  the  plane  angle  may  be  taken  as  the  measure  of  the 
diedral  angle. 


158 


SOLID   GEOMETRY. 


[BK.  VI. 


PROPOSITION  X.     THEOREM. 

369.   A  plane  perpendicular  to  each  of  two  intersecting  planes 
is  perpendicular  to  their  intersection. 


Let  the  planes  PQ  and  RS  be  perpendicular  to  MN. 

To  prove  that  their  intersection  AB  is  perpendicular  to  MN. 

Let  a  perpendicular  be  erected  to  the  plane  MN  at  B. 

Since  B  is  a  point  in  the  plane  RS,  as  RS  is  perpendicular 
to  MN,  the  perpendicular  BA  will  lie  (by  365)  in  RS. 

For  the  same  reason  BA  will  lie  in  PQ. 

Therefore  as  BA  lies  in  both  planes,  it  must  be  in  the  inter- 
section of  those  planes,  or  the  intersection  BA  is  perpendicular 
to  MN.  Q.E.D. 

370.  COR.  1.     If  two  intersecting  planes  are  each  perpen- 
dicular to  a  third  plane,  their  intersection  is  perpendicular  to 
the  third  plane. 

371.  COR.  2.     If  the  planes  PQ  and  RS  include  a  right 
diedral  angle,  the  three  planes  PQ,  RS,  MN,  are  perpendicular 
to  one  another ;  the  intersection  of  any  two  of  these  planes  is 
perpendicular  to  the  third  plane ;  and  the  three  intersections 
are  perpendicular  to  one  another. 

PROPOSITION  XL     THEOREM. 

372.  The  acute  angle  between  a  straight  line  and  its  projection 
on  a  plane  is  the  least  angle  ivhich  the  line  makes  ivith  any  line 
of  the  plane. 


§374.]  POLYEDRAL  ANGLES.  159 

Let  BC  be  the  projection  of  AB  on  the 
plane  3/JV,  and  BD  be  another  line  in  the 
same  plane  passing  through  B. 

To  prove  Z  ABC  <  Z  ABD.  /B 

Take  BD  =  BC,  and  join  AD  and  ^1(7. 

Since  AC  is  the  perpendicular  (by  244) 
to  the  plane,  it  is  shorter  (by  335)  than  any  oblique  line  from 
A  to  the  plane,  or  AC  <  AD. 

In  the  two  triangles  ABC  and  ABD,  the  side  AB  is  com- 
mon, the  side  BD  —  5(7  by  construction,  but  AD  >  AC. 

Therefore  (by  95)  the  greater  angle  lies  opposite  the  greater 
third  side,  or  Z  ABC  <  ABD.  Q.E.D. 

EXERCISE. 

Show  that  a  straight  line  makes  equal  angles  with  parallel  planes. 

POLYEDRAL  ANGLES. 
DEFINITIONS. 

373.  When  three  or  more  planes  meeting  in  a  point  separate 
a  portion  of  space  from  the  rest  of  space,  they  form  or  include 
a  Polyedral  Angle. 

The  common  point  in  which  the  planes  meet  is  the  Vertex  of 
the  angle,  the  intersections  of  the  planes  are  the  Edges,  the 
portion  of  the  planes  between  the  edges  are  the 
Faces,  and  the  plane  angles  formed  by  the  edges 
are  the  Face-angles. 

Thus,  the  point  S  is  the  vertex,  the  straight 
lines  SA,  SB,  etc.,  are  the  edges,  the  planes 
SAB,  SBC,  etc.,  are  the  faces,  and  the  angles 
ASB,  BSC,  etc.,  are  the  face-angles  of  the  polye- 
dral angle  S-ABCD. 

374.  The  edges  of  a  polyedral  angle  may  be  produced  in- 
definitely ;  but  to  represent  the  angle  clearly,  the  edges  and 
faces  are  supposed  to  be  cut  off  by  a  plane,  as  in  the  figure 
above.     The  intersection  of  the  faces  with  this  plane  forms  a 


160  SOLID    GEOMETRY.  [Bit.  VI, 

polygon,  as  ABCD,  which  is  called  the  Base  of  the  polyedral 
angle. 

375.  In  a  polyedral  angle,  each  pair  of  adjacent  faces  forms 
a  diedral  angle,  and  each  pair  of  adjacent  edges  forms  a  face- 
angle.     There  are  as  many  edges  as  faces,  and  therefore  as 
many  diedral  angles  as  faces. 

376.  The  magnitude  of  a  polyedral  angle  depends  only  upon 
the  relative  position  of  its  faces,  and  is  independent  of  their 
extent.     Thus,   by  the   face   SAB  is  not  meant  the  triangle 
SAB,  but  the  indefinite  plane  between  the  edges  SA,  SB  pro- 
duced indefinitely. 

377.  Two  polyedral   angles   are   equal,  when  the  face  and 
diedral  angles  of  the  one  are  respectively  equal  to  the  face 
and  diedral  angles  of  the  other,  and  arranged  in  the  same  order. 

Thus  if  the  face  angles  AOB, 
BOC,  and  CO  A  are  equal  respec- 
tively to  the  face  angles  A'O'B', 
B'O'C',  and  C'O'A',  and  the  die- 
dral  angles  OA,  OB,  and  OC  to 
the  diedral  angles  O'A',  O'B',  and  B  *' 

O'C',  the  triedral  angles  0-ABC  and  O'-A'B'C'  are  equal,  for 
they  can  evidently  be  applied  to  each  other  so  that  their  faces 
shall  coincide. 

378.  Two  polyedral  angles  are  symmetrical  when  the  face 
and  diedral  angles  of  one  are  equal  to  those  of  the  other,  each 
to  each,  but  arranged  in  reverse  order. 

Thus  if  the  face  angles  AOB,  BOC,  and  COA  are  equal 
respectively  to  the  face  angles  A  O'B1,  B'O'C1,  and  C'O'A', 
and  the  diedral  angles  OA,  OB, 
and  OC  to  the  diedral  angles 
O'A,  O'B',  and  O'C",  the  triedral 
angles  0-AB<7and  O'-A'B'C1  are  A/_ 
symmetrical,  for  their  equal  parts 
are  arranged  in  the  reverse  order. 


§  384.] 


POLYEDEAL   ANGLES. 


161 


379.  A  polyedral  angle  of  three  faces  is  called  a   Triedral 
angle,  one  of  four  faces  a  Quadraedral,  etc. 

380.  A  triedral  angle  is  called  Isosceles  if  it  has  two  of  its 
face-angles  equal;  and  Equilateral  if  three  of  its  face-angles  are 
equal. 

381.  Triedral  angles  are  Rectangular,  Bi-rectangular,  or  Tri- 
rectangalar,  according  as  they  have  one,  two,  or  three  right 
diedral  angles. 

382.  A  polyedral  angle  is  Convex,  if  the 
polygon  formed  by  the  intersections  of  a 
plane  with  all  its  faces  be  a  convex  polygon. 

383.  Opposite  or  Vertical  polyedral  angles 
are  those  in  which  the  edges  of  the  one  are 
prolongations  of  the  edges  of  the  other. 

Such  angles  are  symmetrical,  as  0-ABC 
and  0-A'B'C'. 

PROPOSITION  XII.     THEOREM. 

384.  Tlie  sum  of  any  two  face-angles  of  a  triedral  angle  is 
greater  than  the  third. 


If  the  angles  are  equal,  it  is  evident  that  the  sum  of  any  two 
will  be  greater  than  the  third. 

If  unequal,  let  /-AOC  be  greater  than  Z  AOB  or  Z  BOC  in 
the  triedral  Z  0-ABC. 


162  SOLID   GEOMETRY.  [Bit.  VI. 

In  the  plane  AOC  draw  the  line  OD  making  Z  AOD  = 
Z  AOB;  draw  AC  cutting  OD  in  D  and  pass  a  plane  through 
AC  so  that  it  may  cut  off  OB  equal  to  OD. 

Then  the  triangles  OAD  and  OAB  will  be  equal  (by  80), 
having  two  sides  and  the  included  angle  equal  by  construction, 
which  gives  AD  =  AB. 


In  the  triangle  ABC,  AB  +  BC>  AC  (by  77)  ;  subtracting 
the  equals  AB  =  AD,  we  have  BC  >  DC. 

In  the  triangles  BOC  and  DOC,  OB  =  OD,  and  the  side  OC 
is  common,  but  the  third  side  BC  is  greater  than  DC,  therefore 
(by  95)  Z  BOC  >Z  DOC. 

Add  the  equal  angles,  Z  AOB  =  Z  AOD, 


and 

or  Z  .405  -f  ^  -BOO  >  Z  .40(7.  Q.E.D. 

EXERCISES. 

1.  If  a  plane  intersects  all  the  faces  of  a  triedral  angle,  what  kind  of  a 
plane  figure  is  formed  by  the  lines  of  intersection  ?     What  In  the  case  of 
a  tetraedral  angle  ? 

2.  The  sides  of  an  isosceles  triangle  are  everywhere  equally  inclined 
to  any  plane  passing  through  its  base. 


§385.]  POLYEDEAL  ANGLES.  163 

PROPOSITION  XIII.     THEOREM. 

385.    TJie  sum  of  the  face-angles   of  any   convex  polyedral 
angle  is  less  than  four  right  angles. 

O 


Let  0-ABCDE  be  a  convex  polyedral  angle. 

To  prove  that  the  sum  of  the  face  angles  AOB,  BOC,  etc., 
is  less  than  four  right  angles. 

Pass  the  plane  ABODE  intersecting  the  edges  in  A,  B,  C,  D, 
and  E,  and  let  0'  be  any  point  in  this  plane. 

Join  0'  with  A,  B,  Gy,  D,  and  E. 

Since  the  sum  of  any  two  face-angles  at  a  triedral  angle  is 
greater  than  the  third  (by  384), 

Z  OAB  +  Z  OAE  >  Z  EAB, 
also  Z  OB  A  +  Z  OBC  >  Z  ABC,  etc. 

That  is,  the  sum  of  the  base  angles  whose  vertex  is  0  is  greater 
than  the  sum  of  the  base  angles  whose  vertex  is  0'. 

But  the  sum  of  all  the  angles  of  the  triangles  whose  vertex 
is  0  must  be  equal  to  the  sum  of  all  the  angles  of  the  triangles 
whose  vertex  is  0',  since  the  number  of  triangles  in  each  case 
is  the  same,  and  (by  79)  the  value  of  the  angles  of  each  triangle 
is  identical. 

Therefore  the  angles  at  the  vertex  of  the  triangles,  having 
the  common  vertex  0,  is  less  than  the  vertex  angles  at  0',  or 
less  than  four  right  angles.  Q.E.D. 


BOOK   VII. 

POLYEDRONS,  CYLINDERS,  AND  CONES. 


GENERAL  DEFINITIONS. 

386.  A  Polyedron  is  a  solid  bounded  by  planes.     The  Faces 
are  the  bounding  planes,  the  Edges  are  the  intersections  of  its 
faces,  and  the  Vertices  are  the  intersections  of  its  edges. 

387.  The  Diagonal  of  a  polyedron  is  a  straight  line  joining 
any  two  non-adjacent  vertices  not  in  the  same  plane. 

388.  A  polyedron  of  four  faces  is  called  a  Tetraedron;  of 
six  faces,  a  Hexaedron;  of  eight  faces,  an  Octaedron;  of  twelve 
faces,  a  Dodecaedron  ;  of  twenty  faces,  an  Icosaedron. 

389.  A  polyedron  is  called  Convex  when  the  section  made 
by  any  plane  is  a  convex  polygon. 

All  polyedrons  treated  hereafter  will  be  understood  to  be 
convex. 

390.  The  Volume  of  a  solid  is  the  number  which  expresses 
its  ratio  to  some  other  solid  taken  as  a  unit  of  volume.     The 
Unit  of  Volume  is  a  cube  whose  edge  is  a  linear  unit. 

391.  Two  solids  are  Equivalent  when  their  volumes  are  equal. 

PRISMS  AND  PAKALLELOPIPEDS. 

392.  A  Prism  is  a  polyedron  two  of  whose  faces  are  equal 
and  parallel  polygons,  and  the  other  faces  are  parallelograms. 

164 


Die. TIL, §400.]  POLYEDRONS,  CYLINDERS,  AND  CONES.  165 


The  equal  and  parallel  polygons  are  called 
the  Bases  of  the  prism ;  the  parallelograms  are 
the  Lateral  Faces;  the  lateral  faces  taken  together 
form  the  Lateral  or  Convex  Surface;  and  the  in- 
tersections of  the  lateral  faces  are  the  Lateral 
Edges. 

The  lateral  edges  are  parallel  and  equal,  and 
the  area  of  the  lateral  surface  is  called  the  Lateral 
Area. 


393.    The  Altitude  of  a  prism  is  the  perpendicular  distance 
between  its  bases. 


J, 


394-   Prisms  are  Triangular,  Quadrangular,  Pentangular, 
according  as  their  bases  are  triangles,  quadrangles,  pentagons, 
etc. 


395.  A    Right    Prism    is    a    prism    whose 
lateral  edges  are  perpendicular  to  its  bases. 

396.  An   Oblique  Prism  is  a  prism  whose 
lateral  edges  are  oblique  to  its  bases. 


RIGHT    PRISM. 


397.  A  Regular  Prism   is    a   right   prism 

whose  bases  are  regular  polygons,  and  hence  its  lateral  faces 
are  equal  rectangles. 

398.  A  Truncated  Prism  is  a  portion  of  a  prism  included 
between  either  base  and  a  section  inclined  to  the  base  and 
cutting  all  the  lateral  edges. 

399.  A  Right  Section  of  a  prism  is  a  section  perpendicular 
to  its  lateral  edores. 


400.  A  Parallelopiped  is  a  prism  whose 
bases  are  parallelograms ;  therefore  all  the 
faces  are  parallelograms,  and  the  opposite 
faces  are  equal  and  parallel. 


166 


SOLID    GEOMETRY. 


[BK.  VII. 


401.  A  Right  Parallelepiped  is  one  whose  lateral  edges  are 
perpendicular   to   its   bases ;     that   is,   the   lateral   faces   are 
rectangles. 

402.  A  Rectangular  Parallelepiped  is  a  right  parallelepiped 
whose  bases   are  rectangles ;    that  is,  all  the 

faces  are  rectangles. 

Such  a  solid  is  sometimes  called  a  cuboid. 
It  is  contained  between  three  pairs  of  parallel 
planes. 

The  Dimensions  of  a  rectangular  parallelepiped  are  the  three 
edges  which  meet  at  any  vertex. 


~' 


403.  A  Cube  is  a  rectangular  parallelepiped  whose  six  faces 
are  all  squares,  and  edges  consequently  equal. 

404.  Similar  Polyedrons  are  those  which  are  bounded  by 
the  same  number  of  similar  polygons,  similarly  placed. 

Parts  which  are  similarly  placed,  whether  faces,  edges,  or 
angles,  are  called  Homologous. 

405.  A  Cylindrical  Surface  is  a  curved  surface  traced  by 
a  straight  line,   so  moving  as  to 

intersect  a  given  curve  and  always 
be  parallel  to  a  given  straight  line 
not  in  the  curve. 

Thus  if  the  line  EF  moves  so 
as  to  continually  intersect  the 
curve  DC,  and  always  be  parallel 
to  GH,  the  surface  AC  is  a  cylin- 
drical surface. 

406.  The  moving  line  EF  is  the  Generatrix,  the  fixed  curve 
DC  the  Directrix.,  and  EF  in  any  of  its  positions  is  an  Element 
of  the  surface. 

407.  A  General  Cylinder  is  a  solid  bounded  by  a  cylindrical 
surface  and  two  parallel  planes  called  Bases. 


§414.]       POLYEDRONS,    CYLINDERS,   AND   CONES.          167 

408.  The  Lateral  Surface  is  the  curved  surface. 

408  a.  A  plane  which  contains  an  element  of  the  cylinder 
and  does  not  cut  the  surface  is  called  a  tangent  plane,  and  the 
element  contained  by  the  tangent  plane  is  the  element  of 
contact. 

409.  The  Altitude  of  a  cylinder  is  the  per- 
pendicular distance  between  the  bases  or  the 
planes  of  the  bases. 

410.  The   Right    Cylinder   is    the    cylinder 
whose  element  is  perpendicular  to  its  base. 

If  the  base  is  distorted  so  as  to  be  no  longer 
regular,  the  cylinder  is  still   a   right  though   not   a  regular 
cylinder. 

411.  A  Circular  Cylinder  is  one  whose  directrix  is  a  circle. 
NOTE.     Hereafter  the  term  cylinder  is  used  for  circular  cylinder. 

412.  A    right    cylinder    may    be    conceived    as 
formed  by  the  revolution  of  a  rectangle  about  one 
of  its  sides. 

Similar  cylinders  of  revolution  are  generated  by 
similar  rectangles. 

413.  Since  the  base  of  a  cylinder  is  a  polygon  of  an  infinite 
number  of  sides,  the  cylinder  itself  may  be  regarded  as  a  prism 
of  an  infinite  number  of  faces;  that  is,  a  cylinder  is  only  a 
prism  under  this  condition  of  infinite  faces. 

414.  Hence  the  cylinder  will  have  the  properties  of  a  prism, 
and  all  demonstrations  for  prisms  will  include  cylinders  when 
so  stated  in  the  theorem  or  in  the  corollary. 


168  SOLID   GEOMETRY  [BK.  VII. 


PROPOSITION  I.     THEOREM. 

415.    The  lateral  area  of  a  prism  is  equal  to  the  product  of  the 
perimeter  of  a  right  section  by  a  lateral  edge. 


B 

Let  AD'  be  a  prism,  and  FGHIK  a  right  section. 

To  prove  that  the  lateral  area  =  AA'  (FG  +  OH  +  HI,  etc.). 

Since  a  right  section  is  perpendicular  to  the  lateral  edges, 
FG,  GH,  HI,  etc.,  are  altitudes  of  the  parallelograms  which 
form  the  faces  of  the  prism.  Hence, 

area  of  A  ABB'  =  AA  x  FG  (by  251), 
and  area  of  B'BCC'  =  BB'  x  GH,  etc. 

But  (by  392)  the  lateral  edges  are  equal ;  that  is, 

AA'  =  BB'  =  CO',  etc. 
Therefore  the  total  lateral  surface  will  be 

AA  xFG  +  AA  x  GH  +  AA!  x  HI  +  etc., 
or    lateral  surface  AD  =  AA  (FG  +  GH+HI+  etc).     Q.E.D. 

416.  COR.  1.     The  lateral  area  of  a  right  prism  is  equal  to 
the  product  of  the  perimeter  of  its  base  by  its  altitude. 

417.  COR.  2.     The  lateral  area  of  a  cylinder  is  equal  to  the 
perimeter   of  a  right  section  of  the  cylinder  multiplied   by  an 
element. 


§418.]       POLYEDKONS,    CYLINDERS,   AND   CONES.          169 


PROPOSITION  II.     THEOREM. 


418.   An  oblique  prism  is  equivalent  to  a  rfgJ/t  prism  having 


for  its  base  a  right  section  of  the  oblique 
tude  a  lateral  edge  of  the  oblique  prism. 

f'~—^      r' 

•*     I  \  I          •-->./ 


and  for  its  alti- 


Let ABCDE-I  be  an  oblique  prism,  and  A'B'C'D'E'  a  right 
section  of  it. 

Produce  A'Fto  F',  making  A'F'  -.  --  AF,  likewise  B'G'  =  =  BG, 
C'H'=CH,  D'I'  =  DI,  E''K'  =  EK;  then  will  F'-l'  be  a 
plane  (by  350)  parallel  to  A'-D',  which  is  a  right  section ; 
hence  A'B'C'D'E '-I'  will  be  a  right  prism. 

To  prove  that  prism  AI=  prism  A' I'. 

The  truncated  prisms  F-I'  and  A-D'  are  equivalent,  since  the 
faces  FGG'F1  and  ABB' A  are  equal  by  construction,  likewise 
G'GHH'  and  B'BCC',  and  so  with  each  pair  of  faces.  The 
diedral  angles  are  equal,  being  formed  by  a  continuation  of  the 
same  faces  ;  that  is, 

Z.A'A=Z.FF\  ^B'B=ZG'G,  etc. 

Therefore  the  space  occupied  by  A-D'  could  be  exactly  filled 
by  F-I',  or  vice  versa;  that  is,  the  prisms  are  equivalent. 

Hence  if  from  the  entire  solid  .1-7'  we  subtract  the  solid 
A-D',  we  have  left  the  right  prism  A'-F,  and  if  from  the 
same  prism  we  subtract  the  equal  prism  F-I',  we  have  left 
the  oblique  prism  A-L 

Therefore  prism  A-I  =  right  prism  A'-F.  Q.E.D. 


170 


SOLID   GEOMETRY. 


[BK.VII 


PROPOSITION  III.     THEOREM. 

419.    Two  rectangular  parallelepipeds  having  equal  bases  are 
to  each  other  as  their  altitudes. 


C 

A 


f/    A 


Let  P  and  Q  be  two  rectangular  parallelepipeds  having  equal 
bases,  and  let  their  altitudes  AA'  and  BB'  be  commensurable. 

P     AA' 


To  prove  that 


BB' 


Let  AC  be  a  common  measure  of  AA  and  BB',  and  suppose 
it  to  be  contained  4  times  in  AA  and  3  times  in  BB'. 


Then, 


AA' 


(1) 


BB'      3 

At  the  several  points  of  division  of  AA  and  BB'  pass  planes 
perpendicular  to  these  lines. 

Then  the  parallelepiped  P  will  be  divided  into  4  equal  parts, 
of  which  the  parallelepiped  Q  will  contain  3. 

Therefore.  -  =  -•  (2) 

Q     3  w 

From  (1)  and  (2),  we  have 

f=^l 

Q      BB' 

When  the  altitudes  arc  incommensurable,  the  demonstration 
follows  the  method  pursued  in  section  1.73. 


§422.]        POLYEDRONS,    CYLINDERS,    AND    CONES.          171 


420.    SCHOLIUM.     This  theorem  may  also  be  expressed  as 
follows  : 


i'o  rectangular  parallelepipeds  ichich  hace  two  dimensions  i 
common  are  to  each  other  as  their  third  dimensions. 


PROPOSITION  IV.     THEOREM. 

421.    Tico  rectangular  parallelepipeds  having  equal  altitudes 
are  to  each  other  as  their  bases. 

Let  P  and   Q  be  two  rectangular         p 
parallelepipeds   having   the    common 
altitude  c  and  the  rectangles  ab  and 
a'b'  for  bases. 

m  ,i 

lo  prove  that 


Q     a'b' 

Construct  a  third  parallelepiped  R 
which  shall  have  a,  £>',  and  c  for  its 
dimensions. 

Then  since  P  and  R  have  by  con- 
struction two  dimensions  in  common, 
we  have  (from  420) 


For  the  same  reason 


Hence  by  multiplication 


R=aL 
Q     a1' 

P      ab 


Q.E.D. 


422.    SCHOLIUM.     The  theorem  may  also  be  expressed  : 

Tico  rectangular  parallelepipeds  having  one  dimension  in 
common  are  to  each  other  as  the  products  of  their  other  two 
dimensions. 


172 


SOLID   GEOMETRY. 


[BK.  VII. 


PROPOSITION  V.     THEOREM. 

423.   Any  two  rectangular  parallelopipeds  are  to  each  other  as 
the  products  of  their  three  dimensions. 


I 


a 


Let  P  and  Q  be  two  rectangular  parallelopipeds  having  the 
dimensions  a,  b,  c,  and  a',  &',  c',  respectively. 

To  prove  that  -  =  ax  b  Xc  . 

Q     a1  x  b'  x  c' 

Construct  a  third  parallelepiped  having  the  dimensions  a', 
6',  and  c. 

Then  since  P  and  R  have  the  dimension  c  in  common,  we 
have  (by  422) 

P      axb 


Ra'  x  &' 

Again,  R  and  Q  have  the  two  dimensions  a'  and  b'  .common, 
hence  (by  420)  we  have 


Multiplying  these  equal  ratios,  it  gives 

P       a  x  b  x  c 
Q      a'  x  b'  x  c' 


Q.E.D. 


424.  COR.  1.  If  a'  =  b'  =  c'  =  1,  then  Q  will  be  the  unit  of 
volume,  and  the  above  proportion  becomes  P  —  a  x  b  x  c,  or 
the  product  of  its  three  dimensions. 


§427.]        POLYEDEONS,    CYLINDERS,    AND    CONES.  173 

425.  COR.  2.     Since  a  x  b  gives  the  area  of  the  base  (from 
248),  we  have  the  volume  of  a  rectangular  parallelepiped  equal 
to  the  product  of  its  base  by  its  altitude. 

426.  COR.  3.     If  a  =  b  =  c,  then  (from  424)  P=  a  x  a  x  a  =•- a3 ; 
that  is,  the  volume  of  a  cube  (403)  is  the  cube  of  its  edge. 

EXERCISES, 

1.  Show  that  the  diagonals  of  a  parallelepiped  bisect  each  other. 

2.  Show  that  the  square  of  a  diagonal  of  a  rectangular  parallelepiped 
is  equal  to  the  sum  of  the  squares  of  the  three  edges  meeting  at  any 
vertex. 

PROPOSITION  VI.     THEOREM. 

427.  The  volume  of  any  parattelopiped  is  equal  to  the  product 
of  its  base  and  altitude. 


Let  P  be  any  parallelepiped  with  the  base  B  and  altitude  h. 
To  prove  that  vol.  P  =  B  x  h. 

Extend  the  lines  CD  and  EF  and  also  the  corresponding 
lines  of  the  base,  and  construct  thereon  the  indefinite  prism  P. 
Cut  from  this  the  right  prism  Q,  whose  altitude  HK  is  equal 
to  the  lateral  edge  CD. 


174 


SOLID    GEOMETRY. 


[BK.  VII. 


Then  (from  418)  the  oblique  prism  P  =  right  prism  Q. 

Extend  the  lines  GH  and  IK  and  also  the  corresponding 
lines  of  the  base  of  Q,  and  construct  thereon  the  indefinite 
prism  Q'.  From  this  cut  the  rectangular  prism  R,  having  its 
altitude  equal  to  the  lateral  edge  of  Q  and  base  B". 


Then  (from  423)  Q  -  -  R. 

But  it  was  shown  that  P  =  Q,  therefore  P  =  R. 

Now  R  is  a  rectangular  parallelepiped  since  its  faces  are 
perpendicular  to  each  other,  that  gives  R  =  B"  x  h. 

But  the  parallelograms  B  and  B'  are  equal,  being  between 
the  same  parallels ;  likewise  B'  =  B"  for  the  same  reason, 
therefore  B  =  B". 

Hence  R  —  B  x  h.  Q.E.D. 


EXERCISES. 

1.  Show  that  in  any  parallelepiped  the  sum  of  the  squares  of  the 
twelve  edges  is  equal  to  the  sum  of  the  squares  of  its  four  diagonals. 

2.  Find  the  length  of  the  diagonal  of  a  rectangular  parallelepiped 
whose  dimensions  are  3,  4,  and  5. 

3.  Find  the  volume  of  a  rectangular  parallelepiped  whose  surface  is 
932  and  whose  base  is  4  by  12. 

4.  Find  the  side  of  a  cube  which  contains  as  much  as  a  rectangular 
parallelepiped  16  feet  long,  4  feet  wide,  and  3  feet  high. 


§429.]       POLYEDRONS,    CYLINDERS,   AND   CONES.          175 


PROPOSITION  VII.     THEOREM. 

428.    The  volume  of  a  triangular  prism  is  equal  to  the  product 
of  its  base  and  altitude. 


Let  AE  be  the  altitude  of  the  triangular  prism  ABC-C'. 
To  prove  that 

volume  ABC-C'  =  ABC  x  AE. 

Construct  the  parallelepiped  ABCD-D'  having  its  edges 
equal  and  parallel  to  AB,  BC,  and  BB'. 

Since  the  diagonal  AC  divides  the  parallelogram  ABCD  into 
two  equal  parts,  the  two  prisms  ABC-C'  and  ADC-D',  each 
being  equivalent  to  a  right  prism  of  the  same  altitude  and 
equal  right  section  are  equivalent. 

But  the  parallelepiped  (by  427)  is  equal  to  the  product  of 
its  base  by  its  altitude. 

Therefore  the  half  parallelepiped  is  equal  to  the  product  of 
the  half  base  by  its  altitude ;  that  is, 


volume  ABC-C'  =  ABC  x  AE. 


Q.E.D. 


429.  COR.  Since  any  prism  can  be  divided  into  triangular 
prisms  by  diagonal  planes,  each  prism  being  equal  to  the  product 
of  its  base  by  its  altitude,  it  follows  that  the  volume  of  any  prism 
is  equal  to  the  product  of  its  base  and  altitude. 


176 


SOLID   GEOMETRY. 


[BK.  VII. 


PROPOSITION  VIII.     THEOREM. 

430.   Similar  triangular  prisms  are  to  each  other  as  the  cubes 
of  their  homologous  edges. 


C' 


w 

/  '"^    ' 

<—fi-l. 

1 

B' 


Let  CBD-P  and  C'B'D'-P'  be  two  similar  triangular  prisms, 
and  let  BC  and  B'C'  be  any  two  homologous  edges. 
To  prove  that 

CBD-P :  C'B'D'-P'  =  BC3 :  B^1'3. 

Since  the  homologous  triedral  angles  B  and  B'  are  equal,  and 
the  faces  which  bound  them  are  (by  404)  similar,  these  triedral 
angles  may  be  applied,  one  to  the  other,  so  that  the  angle 
C'B'D'  will  coincide  with  CBD,  with  the  edge  B'A'  on  BA. 

In  this  case  the  prism  C'B'D'-P'  will  take  the  position  of 
cBd-p. 

From  A  draw  AH  perpendicular  to  the  common  base  of  the 
prisms;  then  the  plane  BAH  is  (by  365)  perpendicular  to  the 
plane  of  the  base. 

From  a  draw  ah  likewise  in  the  plane  BAH  perpendicular  to 
the  intersector  BH}  and  it  will  (by  364)  be  perpendicular  to  the 
plane  of  the  base. 

Since  the  bases  BCD  and  Bed  are  similar  (by  262), 

CBD:cBd=CB2:^B2      .     .     .     .     .     (a) 

In  the  similar  triangles  ABH  and  aBh  (by  218), 

AH:ah  =  AB:aB. 


§434.]       POLYEDROXS,    CYLINDERS,    A XI)    CONES.          177 

111  the  similar  parallelograms  AC  and  ac  (by  223), 

AB:aB  =  BC:Bc; 
therefore  (by  28) 

AH:ah=CB:cB (b) 

Multiplying  (a)  by  (&),  we  have 


CBD  xAH:  cBd  x  ah  =  CB  :  cB. 

But  (by  428)   CBD  x  AH  is  the  volume  of  CBD-P,  and 
cBd  x  ah  is  the  volume  of  C'B'D'-P',  and  cB  =  C"£'. 


Therefore     0££>-P  :  C'B'D'-P'  =  CB"  :  CPSr,  Q.E.D. 


431.  COR.  1.    ^4n?/  foro  similar  prisms  are  to  each  other  as  the 
cubes  of  their  homologous  edges. 

For,  since  the  prisms  are  similar,  their  bases  are  similar  poly- 
gons (by  4Q£);  and  these  similar  polygons  may  each  be  divided 
into  the  same  number  of  similar  triangles,  similarly  placed 
(by  121);  therefore,  each  prism  may  be  divided  into  the  same 
number  of  triangular  prisms,  having  their  faces  similar  and  like 
placed  ;  consequently,  the  triangular  prisms  are  similar  (by  404). 
But  these,  triangular  prisms  are  to  each  other  as  the  cubes  of  their 
homologous  edges,  and  being  like  parts  of  the  polygonal  prisms, 
the  polygonal  prisms  themselves  are  to  each  other  as  the  cubes  of 
their  homologous  edges. 

432.  COR.  2.   Similar  prisms  are  to  each  other  as  the  cubes  of 
their  altitudes,  or  as  the  cubes  of  any  other  homologous  lines. 

433.  COR.  3.    Since  the  cylinder  is  the  limit  of  a  prism  of  in- 
finite number  of  sides,  it  follows  that: 

The  volume  of  a  cylinder  is  equal  to  the  product  of  its  base  and 
altitude. 

434.  COR.  4.    The  volumes  of  two  prisms  (cylinders)  are  to 
each  other  as  the  product  of  their  bases  and  altitudes:   prisms 


178  SOLID   GEOMETRY.  [Bit.  VII. 

(cylinders)  having  equivalent  bases  are  to  each  other  as  their 
altitudes:  prisms  (cylinders)  having  equal  altitudes  are  to  each 
other  as  their  bases:  prisms  (cylinders)  having  equivalent  bases 
and  equal  altitudes  are  equivalent. 

EXERCISES. 

Find  the  lateral  area  and  volume,  when  right : 

1.  Of  a  triangular  prism,  each  side  of  whose  base  is  3,  and  whose  alti- 
tude is  8. 

2.  Of  a  regular  hexagonal  prism,  each  side  of  whose  base  is  2,  and 
whose  altitude  is  12. 

3.  Of  a  triangular  prism  whose  altitude  is  18  and  the  sides  of  the  base 
are  6,  8,  and  10. 

PYRAMIDS. 

435.  A  Pyramid  is  a  polyedron,  one  of  whose  faces  is  a 
polygon,  and  whose  other  faces  are  triangles  having  a  common 
vertex  without  the  base  and  the   sides  of   the   polygon  for 
bases.  V 

436.  The  polygon  ABODE  is  the 
Base  of  the  pyramid,  the  point   V 
the    Vertex,    VBC,    VCD,    etc.,    the 
Lateral,  or  Convex  Surface,  VC,  VB, 
etc.,  the  Lateral  edges,  and  the  area 
of  the  lateral  -  surface  is  called  the 
Lateral  Area. 

C  B 

437.  The  Altitude  of  a  pyramid  is  the  perpendicular  dis- 
tance from  the  vertex  to  the  plane  of  the  base. 

438.  A  pyramid  is  called    Triangular,  Quadrangular,  Pen- 
tagonal, etc.,  according  as  its  base  is  a  triangle,  quadrilateral, 
pentagon,  etc. 


§445.]       POLYEDRONS,    CYLINDERS,   AND   CONES.          179 

439.  A  triangular  pyramid  has  but  four  faces,  and  is  called 
a  Tetmedron;  any  one  of  its  faces  can  be  taken  for  its  base. 

440.  A  Regular  Pyramid  is  one  whose  base  is  a  regular 
polygon,  the  centre  of  which  coincides  with  the  foot  of  the 
perpendicular  let  fall  upon  it  from  the  vertex.     The  lateral 
edges   of  a  regular  pyramid  are   (by  334)   equal,   hence   the 
lateral  faces  are  equal  isosceles  triangles. 

441.  The  Slant  Height  of  a  regular  pyramid  is  the  altitude 
of  any  one  of  its  lateral  faces ;  that  is,  the  straight  line  drawn 
from  the  vertex  of  the  pyramid  to  the  middle  point  of  any 
side  of  the  base. 

442.  A   Truncated  Pyramid  is  the   portion   of  a   pyramid 
included  between  its  base  and  a  plane  cutting  all  the  lateral 
edges. 

443.  A  Frustum  of  a  pyramid  is   a  trun- 
cated pyramid  whose  bases  are  parallel. 

The  Altitude  of  a  frustum  is  the  perpendicu- 
lar distance  between  the  planes  of  its  bases. 

444.  The  lateral  faces  of  a  frustum  of  a  regular  pyramid 
are  equal  trapezoids. 

The  Slant  Height  of  a  frustum  of  a  regular  pyramid  is  the 
altitude  of  any  one  of  its  lateral  faces. 

445.  A  Conical  Surface  is  traced  by  a 
straight  line  so  moving  that  it  always  in- 
tersects a  given  curve  and  passes  through 
a  given  point. 

Thus  the  straight  line  BB'  continu- 
ally intersects  the  curve  ABC  and  passes 
through  the  point  0,  tracing  the  conical 
surface  ABC-0-A'B'C'. 


180 


SOLID   GEOMETRY. 


[BK.  VII. 


446.  The  straight  line  BB'  is  the  Generatrix,  the  curve 
ABC  the  Directrix,  0  the  Vertex,  and  0-ABC,  0-A'B'C'  are 
the  two  Nappes,  and  OB  is  an  Element. 


447.  A  CWe  is  a  solid  bounded  by  a  conical  surface  and 
a   plane  which  cuts  all  of  the  elements  of  the  surface,  as 
0-ABC. 

448.  This  plane  is  called  the  Base,  and  the  perpendicular 
from  the  vertex  to  the  plane  of  the  base  is  the  Altitude. 

449.  A  Circular  Cone  is  one  whose  base  is  a  circle. 
NOTE.     Hereafter  Cone  will  be  used  for  circular  cone. 


450.  A  Right  Cone  is  a  cone  in  which  the 
perpendicular  let  fall  from  the  vertex  meets 
the  base  in  its  centre ;  it  is  also  called  a  cone 
of  revolution,  since  it  can  be  formed  by  re- 
volving a  right  triangle  about  one  of  its 
shorter  sides,  as  V-ABC. 


451.  Since  the  cone  has  a  circular  base  which  is  the  limit 
of  a  polygonal  base,  a  cone  may  be  regarded  as  a  pyramid  of 
an  infinite  number  of  faces,  hence  the  cone  will  have,  in  gen- 
eral, the  properties  of  a  pyramid,  and  all  demonstrations  for 
pyramids  will  include  cones  when  so  stated  in  the  theorem  or 
in  the  corollary. 

452.  A   Truncated  Cone  is  the  portion  of  a  cone   included 
between  its  base  and  another  plane  cutting  all  its  elements. 

453.  A  Frustum  of  a  cone  is  a  truncated 
cone  whose  cutting  planes  or  bases  are  par- 
allel. 

The  Altitude  of  a  frustum  is  the  perpen- 
dicular distance  between  the  planes  of  its 
bases. 


§454.]       POLYEDRONS,    CYLINDERS,   AND   CONES. 


181 


PROPOSITION  IX.     THEOREM. 

454.    If  a  pyramid  is  cut  by  a  plane  parallel  to  its  base : 

(1)  The  edges  and  the  altitude  are  divided 
proportionally. 

(2)  The  section  is  a  polygon  similar  to 
the  base. 

Let  V-ABCDE  be  a  pyramid   cut   by 
the  plane  abcde  parallel  to  the  base. 

1.    To  prove  A' 

Va      Vb  Vo 


VA 


VO 


Suppose  a  plane  to  pass  through  V  parallel  also  to  the  base ; 
then  (by  355) 

Va      Vb  Vo 


VA"VB 


VO 


2.  To  prove  that  the  section  abcde  is  similar  to  the  base 
ABCDE. 

Since  ab  is  parallel  to  AB  and  be  parallel  to  BC,  then  (by 
353)  Z  abc  =  Z  ABC-,  likewise  Z  bed  =  Z  BCD,  etc. 

Again,  ab  and  AB  being  parallel,  we  have  (by  218) 

ab       Vb 


also 

hence 

similarly 


AB     VB' 

be  =  Vb  . 
BC  ~~  VB  ' 

ab        be 


_  =       ,  etc. 


BC     CD 


Therefore  the  polygons  abcde  and  ABCDE  are  mutually 
equiangular  and  have  their  homologous  sides  proportional; 
hence  (by  223)  they  are  similar.  Q.E.D. 


182 


SOLID   GEOMETRY. 


[Biv.  VII. 


455.   COR.  1.     Since  abcde  and  ABCDE  are  similar  poly- 
gons, we  have  (from  264) 

2  T!^2 


abcae          uu         y  u        \  u 
= = =  -—^—  i   Lnaij  is. 

ABCDE     Aff      VB2     VO 


The  areas  of  parallel  sections  of  a  pyramid  are  'proportional  to 
the  squares  of  their  oblique  or  vertical  distances  from  the  vertex. 


B  C 

456.  COR.  2.     In  two  pyramids  of  equal  altitudes  and  equiva- 
lent bases,  sections  made  by  planes  parallel  to  their  bases  and  at 
equal  distances  from  their  vertices  are  equivalent. 

457.  COR.  3.     The  section  of  a  circular  cone  made  by  a  plane 
parallel  to  the  base  is  a  circle. 

PROPOSITION  X.     THEOREM. 

458.  Tfie  lateral  area  of  a  regular  pyramid  is  equal  to  the 
perimeter  of  its  base  multiplied  by  one-half  its  slant  height. 


B  C 

Let  V-ABCDE  be  a  regular  pyramid,  and   VH  the   slant 
height. 


§461.]       POLYEDRONS,    CYLINDERS,   AND   CONES.          183 

To  prove  that 

lateral  area  V-  ABODE  =  (AB  +  BC  +  etc.)  x  \  VH. 

The  lateral  area  of  the  pyramid  is  equal  to  the  sum  of  the 
areas  of  the  triangles  VAB,  VBC,  etc. 

But  (by  254)        area  VAB  =  J-  AB  x  VH\ 
likewise  area  of  VBC  =  \  BC  x  VH,  etc. 

Therefore 

lateral  area  V-ABCDE  =  1  ABxVH+%  BCx  Fff+etc., 
or  =  i  (AB  +  BC  +  etc.)  x  VH.       Q.E.D. 

459.  COR.  1.    The  lateral  area  of  a  frustum  of  a  regular  pyra- 
mid is  equal  to  one-half  the  sum  of  the  perimeters  of  its  bases 
multiplied  by  its  slant  height. 

460.  COR.  2.      The  lateral  area  of  a  cone  of  revolution  is  equal 
to  the  circumference  of  its  base  multiplied  by  one-half  its  slant 
height. 

461.  COR.  3.     The  lateral  area   of  a  frustum  of  a  cone  of 
revolution  is  equal  to  one-half  the  sum  of  the  circumferences  of 
its  bases  multiplied  by  its  slant  height. 

If  R  and  R'  denote  the  radii  of  the  lower  and  upper  bases  of 
the  frustum  of  a  cone  of  revolution,  and  L  the  slant  height, 
then 

lateral  area  =  \  r  [2  irR  +  2  7rR'~\  x  L 

=  TT  (72  +  R')  x  L. 


EXERCISES. 

1.  The  radius  of  the  lower  base  of  the  frustum  of  a  cone  of  revolution 
is  12,  the  radius  of   the  upper  base  is  0  and  the  altitude  8.     Find   the 
lateral  area. 

2.  In  the  above  what  is  the  lateral  area  of  the  cone  that  was  cut  off  to 
form  this  frustum  ? 


184 


SOLID   GEOMETRY. 


[BK.  VII. 


PROPOSITION  XI.     THEOREM. 

462.    Two  triangular  pyramids  having  equivalent  bases  and 
equal  altitudes  are  equivalent. 

V  y' 


Let  V-ABC  and  V'-A'B'C'  be  two  triangular  pyramids 
having  equivalent  bases  ABC  and  A'B'C1  and  a  common 
altitude. 

To  prove  vol.  V-ABC  =  vol.  V'-A'B'C'. 

Divide  the  common  altitude  into  any  number  of  equal  parts, 
and  through  these  points  of  division  pass  planes  parallel  to 
the  plane  of  the  bases,  say  DEF,  D'E'F';  GHI,  G'HT;  etc. 

Upon  DEF  construct  the  prism  DEF-M,  and  on  D'E'F1  the 
prism  D'E'F'-M'.  These  prisms  are  (by  434)  equivalent. 

Likewise,  prism  upon  GHI  is  equivalent  to  the  prism  upon 
G'H'I',  and  so  on. 

Therefore  the  sum  of  the  prisms  in  V-ABC  is  equivalent 
to  the  sum  of  the  prisms  in  V'-A'B'C'. 

Now  let  the  number  of  divisions  be  indefinitely  increased, 
then  the  sum  of  the  prisms  in  V-ABC  will  approach  the 
pyramid  V-ABC  as  its  limit,  and  the  sum  of  the  prisms  in 
V'-A'B'C1  will  approach  the  pyramid  V- A'B'C1  as  its  limit. 

Therefore,  since  the  sums  of  these  prisms  are  always  equiva- 
lent, their  limits  are  equivalent,  or 

vol.  V-ABC  =  vol.  V'-A'B'C'.  Q.E.D. 


§464.]       POLTEDRONS,    CYLINDERS,   AND   CONES.         185 

463.  Since  any  pyramid  can  be  divided  into  triangular  pyra- 
mids by  passing  planes  through  the  vertex  and  the  diagonals 
of  the  base,  it  follows  that  any  two  pyramids  of  equal  altitudes 
and  equivalent  bases  are  equivalent. 


PROPOSITION  XII.     THEOREM. 

464.    The  volume  of  a  triangular  pyr umid  is  equal  to  one-third 
of  the  product  of  its  base  and  altitude. 


Let  V-ABC  be  a  triangular  pyramid  with  h  for  its  alti- 
tude and  ABC  its  base. 

To  prove  that  vol.  V-ABC  =  \  h  x  ABC. 

Upon  the  base  ABC  construct  the  prism  ABC-D,  having 
its  lateral  edges  parallel  to  VB,  and  its  altitude  equal  to  h,  or 
that  of  the  pyramid. 

Draw  AD,  the  diagonal,  and  it  will  divide  (by  107)  the  paral- 
lelogram EACD  into  two  equal  triangles,  and  through  AD  and 
V  conceive  a  plane  to  pass. 

Then  the  prism  will  be  divided  into  three  triangular  pyra- 
mids, V-ABC,  A-VED,  and  A-VCD. 

V-ABC  =  A-VED,  having  equivalent  bases  and  equal  alti- 
tudes. 

V-ABC  can  be  regarded  as  having  A  for  its  vertex   and 


186  SOLID   GEOMETRY.  [Bit.  VII. 

VBC  for  its  base  ;  then  A-VBC  =  A-VCD,  having  a  common 
vertex  and  equal  bases  (by  107). 

Therefore  the  three  triangular  pyramids  are  equivalent,  and 
each,  say  V-ABC,  will  be  one-third  of  the  triangular  prism. 

But  the  volume  of  the  prism  is  (by  428)  equal  to  the  prod- 
uct of  its  base  by  its  altitude,  then  the  volume  of  V-ABC 
=  i  h  x  ABC.  Q.E.D. 

465.  COB.   1.     /Since  any  pyramid  can   be  divided  into  tri- 
angular  pyramids  by  passing  planes  through  the  vertex  and  the 
diagonals  of  its  base,  it  follows  that  the  volume  of  any  pyramid 
is  equal  to  one-third  the  product  of  its  base  and  altitude. 

466.  Con.  2.     The  volume  of  any  cone  is  equal  to  the  product 
of  one-third  of  its  base  by  its  altitude. 

467.  COR.  3.     The  volumes  of  two  pyramids  (cones)  are  to 
each  other  as  the  product  of  their  bases  and  altitudes:    having 
equivalent  bases  they  are  to  each  other  as  their  altitudes:  having 
the  same  altitude  they  are  to  each  other  as  their  bases:  having 
equivalent  bases  and  equal  altitudes  they  are  equivalent. 

468.  COR.  4.     If  a  triangle  and  a  rectangle  having  the  same 
base  and  equal  altitudes  be  revolved  about  the  common  base  as 
an  axis,  the  volume  generated  by  the  triangle  ivill  be  one-third  that 
generated  by  the  rectangle. 

PROPOSITION  XIII.     THEOREM. 

469.  A  frustum  of  a  triangular  pyramid  is  equivalent  to  the 
sum  of  three  pyramids,  having  for  their  common  altitude  the  alti- 
tude of  the  frustum,  and  for  their  bases  the  lower  base,  the  upper 
base,  and  a  mean  proportional  between  the  bases,  of  the  frustum. 

Let  AF  be  a  frustum  of  a  triangular  pyramid. 
Denote  the  area  of  the  lower  base  by  B,  the  area  of  the 
upper  base  by  b,  and  the  altitude  by  h. 


§469.]       POLYEDRONS,    CYLINDERS,   AND   CONES. 


187 


To  prove  that 
vol. 


<b     (200) 
=  i  h  x  (B  +  b  +V£x  6). 

Pass  a  plane  through  the  points  A,  C,  and  E,  and  another 
through  the  points  C,  D,  and  E,  dividing  the  frustum  into 
three  triangular  pyramids,  E-ABC,  C-DEF,  and  E-ACD. 


Let  these  pyramids  be  denoted  by  P,  Q,  and  R,  respectively. 
The  pyramid  P  has  for  its  altitude  the  altitude  h  of  the 
frustum,  and  for  its  base  the  lower  base  B  of  the  frustum. 

Hence  (by  464)  P  =  1  h  X  B.  (1) 

And  the  pyramid  Q  has  for  its  altitude  the  altitude  of  the 
frustum,  and  for  its  base  the  upper  base  b  of  the  frustum. 

Hence  Q  =  J-  h  x  b.  (2) 

Now  the  pyramids  E-ABC  and  E-ACD  may  be  regarded 
as  having  the  common  vertex  C}  and  their  bases  AEB  and 
AED  in  the  same  plane. 

Then  they  have  the  same  altitude,  and  are  to  each  other  (by 
467)  as  their  bases. 

But  the  triangles  AEB  and  AED  have  for  their  common 
altitude  the  altitude  of  the  trapezoid  ABED,  and  are  to  each 
other  as  their  bases  AB  and  DE  (by  256). 

P      AEB      AB 


Therefore 


AED     DE 


(3) 


188  SOLID   GEOMETRY.  [BK.  VII. 

Again,  the  pyramids  E-ACD  and  E-CDF  have  the  com- 
mon vertex  E,  and  their  bases  ACD  and  CDF  in  the  same 
plane. 

Then  they  have  the  same  altitude,  and  are  to  each  other  as 
their  bases. 

But  the  triangles  ACD  and  CDF  have  for  their  common 
altitude  the  altitude  of  the  trapezoid  ACFD,  and  are  to  each 
other  as  their  bases  AC  and  DF. 

Therefore  E  =  ACD  =  AV  ^ 

Q      CDF     DF 

Now  since  the  section  DEF  is  similar  to  the  base  ABC  (by 
454), 

AC 


DF     DE 

Whence  from  (3)  and  (4)  (by  28), 


Substituting  in  this  equation  the  values  of  P  and  Q  from 
(1)  and  (2), 


Whence,  R  =  1  h  x  V-B  x  b. 

Therefore  vol.  AF  =  P  +  Q  +  R 


x  6).      Q.E.D. 

470.  COR.  1.    By  the  same  reasoning  as  in  465  we  may  con- 
clude that :  A  frustum  of  any  pyramid  is  equivalent  to  the  sum 
of  three  pyramids,  having  the  same  altitude  as  the  frustum,  and 
whose  bases  are  the  lower  base,  the  upper  base,  and  a  mean  pro- 
portional between  the  bases,  of  the  frustum. 

471.  COR.  2.    The  volume  of  a  frustum  of  any  cone  is  equal  to 
the  sum  of  the  volumes  of  three  cones,  whose  common  altitude  is 
the  altitude  of  the  frustum,  and  ivhose  bases  are  the  lower  base, 


§472.]       POLYEDRON  S,    CYLINDERS,   AND   CONES.          189 

the  upper  base,  and  a  mean  proportional  between  the  bases  of  the 
frustum. 

If  R  and  R'  denote  the  radii  of  the  lower  and  upper  bases 
of  the  frustum,  and  h  the  altitude,  then  B  =  irR2,  and  b  =  7rR12, 
hence  V-B  x  b  =  irRR'. 

Therefore       vol.  =  1  h  X  IT \_R-  +  R'2  +  RR'~\. 

EXERCISES 

Find  the  lateral  edge,  lateral  area,  and  volume : 

1.  Of  a  regular  triangular  pyramid,  each  side  of  whose  base  is  6,  and 
whose  altitude  is  5. 

2.  Of  a  regular  quadrangular  pyramid,  each  side  of  whose  base  is  16, 
and  whose  altitude  is  18. 

3.  Of  a  regular  hexagonal  pyramid,  each  side  of  whose  base  is  2,  and 
whose  altitude  is  14. 

4.  Of  a  frustum  of  a  regular  hexagonal  pyramid,  the  sides  of  whose 
bases  are  8  and  3,  and  whose  altitude  is  6. 

5.  What  is  the  volume  of  a  frustum  of  a  regular  triangular  pyramid, 
the  sides  of  whose  bases  are  8  and  6,  and  whose  lateral  edge  is  7  ? 

6.  Show  that  the  number  of  plane  angles  at  the  vertices  of  a  poly- 
edron  is  an  even  number. 

7.  The  sum  of  the  face-angles  of  any  polyedron  is  equal  to  four  right 
angles  taken  as  many  times  as  the  polyedron  has  vertices  less  two. 

8.  The  base  of  a  pyramid  is  regular,  if  its  faces  are  equal  isosceles 
triangles. 

9.  Find  the  difference  between  the  volume  of  the  frustum  of  a  pyra- 
mid and  the  volume  of  a  prism  of  the  same  altitude  whose  base  is  a  sec- 
tion of  the  frustum  parallel  to  its  bases  and  equidistant  from  them. 

REGULAR  POLYEDRONS. 

472.  A  Regular  Pobiedron  is  one  whose  faces  are  all  equal 
regular  polygons,  and  whose  polyedral  angles  are  all  equal. 


190  SOLID   GEOMETEY.  [Bit.  VII. 


PROPOSITION  XIV.     THEOREM. 

473.  There  can  be  only  five  regular  convex  polyedrons. 

PROOF.  At  least  three  faces  are  necessary  to  form  a  poly- 
edral  angle,  and  the  sum  of  its  face-angles  must  be  less  than 
360°  (by  385). 

1.  Because  the  angle  of  an  equilateral  triangle  is  60°,  each 
convex  polyedral  angle  may  have  3,  4,  or  5  equilateral  tri- 
angles.    It  cannot  have  6  faces,  because  the  sum  of  6  such 
angles  is  360°,  reaching  the  limit.     Therefore  no  more  than 
three  regular  convex  polyedrons  can  be  formed  with  equilateral 
triangles;  the  tetraedron,  octaedron,  and  icosaedron. 

2.  Because  the  angle  of  a  square  is  90°,  each  convex  poly- 
edral angle  may  have  3  squares.     It  cannot  have  4  squares, 
because  the  sum  of  4  such  angles  is  360°.     Therefore  only  one 
regular  convex  polyedron  can  be  formed  with  squares ;    the 
hexaedron,  or  cube. 

3.  Because  the  angle  of  a  regular  pentagon  is  108°,  each 
convex  polyedral  angle  may  have  3  regular  pentagons.      It 
cannot  have  4  faces,  because  the  sum  of  4  such  angles  is  432°. 
Therefore  only  one  regular  convex  polyedron  can  be  formed  of 
regular  pentagons ;  the  dodecaedron. 

Because  the  angle  of  a  regular  hexagon  is  120°,  and  the 
angle  of  every  regular  polygon  of  more  than  6  sides  is  yet 
greater  than  120°,  therefore  there  can  be  110  regular  convex 
polyedron  formed  of  regular  hexagons  or  of  any  regular 
polygons  of  more  than  6  sides. 

Therefore  there  can  be  only  five  regular  convex  polyedrons. 

Q.E.D. 

474.  SCHOLIUM.     Models  of  the  regular  polyedrons  may  be 
easily  constructed  as  follows  : 

Draw  the  following  diagrams  on  cardboard,  and  cut  them 
out.  Then  cut  halfway  through  the  board  in  the  dividing 


§474.]       POLYED11ONS,    CYLINDERS,    AND    CONES.          191 


lines,  and  bring  the  edges  together  so  as  to  form  the  respective 
polyedrons. 


TETRAEDRON 


HEXAEDRON 


OCTAEDRON 


DODECAEDRON 


ICOSAEORON 


BOOK   VIII. 


THE   SPHERE. 


DEFINITIONS. 

475.  A  Sphere  is  a  solid  bounded  by  a  surface,  all  points 
of  which  are  equally  distant  from  a  point  within  called  the 
centre.  A  sphere  may  be  generated  by  the  revolution  of  a 
semicircle  about  its  diameter  as  an  axis. 


476.  A  Radius   of   a   sphere   is   the 
distance  from  its  centre  to  any  point  in 
the  surface.     All  the  radii  of  a  sphere 
are  equal. 

477.  A  Diameter  of  a  sphere  is  any 
straight  line  passing  through  the  centre 
and  having  its  extremities  in  the  sur- 
face of  the  sphere.     All  the  diameters 

of  a  sphere  are  equal,  since  each  is  equal  to  twice  the  radius. 

478.  A  Section  of  a  sphere  is  a  plane  figure  whose  boundary 
is  the  intersection  of  its  plane  with  the  surface  of  the  sphere. 

479.  Every  section  of  a  sphere  made  by  a  plane  is  a  circle 
(see  334). 

When  the  plane  passes  through  the  centre,  the  section  is 
called  a  Great  Circle. 

480.  Every  great  circle  plane  bisects  the  sphere. 

481.  Any  two  great  circles  bisect  each  other. 

192 


BK.  VIII. ,  §  487.]  THE   SPHERE.  193 

482.  An  Axis  of  a  circle  of  a  sphere  is  the  diameter  of  the 
sphere  perpendicular  to  the  circle;  and  the  extremities  of  the 
axis  are  the  Poles  of  the  circle. 

483.  All  points  in  the  circumference  of  a  circle  of  a  sphere 
are  equally  distant  from  each  of  its  poles,  the  distance  being- 
measured  along  the  arcs  of  a  great  circle  (see  334). 

484.  A  straight  line  or  a  plane  is  said  to  be  tangent  to  a 
sphere  when  it  has  but  one  point  in  common  with  the  surface 
of  the  sphere. 

The  common  point  is  called  the  Point  of  Contact,  or  Point  of 
Tangency. 

485.  A  plane  perpendicular  to  a  radius  at  its  extremity  is 
tangent  to  the  sphere.     (See  150.) 

486.  A  great  circle  can  be  passed  through  any  two  points 
on  a  sphere,  since  a  plane  can  be  made  to  pass 

through  these  points  and  the  centre,  thus  in- 
tersecting the  surface  of  the  sphere  in  a  great 
circle. 

By  distance  between  two  points  is  meant 
the  shorter  arc  of  the  great  circle  passing 
through  them,  as  CD. 


PROPOSITION  I.     THEOREM. 

487.  If  a  point  on  the  surface  of  a  sphere  lies  at  a  quadrant's 
distance  from  each  of  two  points  in  the  arc  of  a  great  circle,  it  is 
the  pole  of  that  arc. 

Let  the  point  P  be  a  quadrant's  distance  from  each  of  the 
points  A  and  B-,  that  is,  the  arc  joining  P  and  A  is  one-fourth 
of  the  arc  of  a  great  circle. 

To  prove  that  P  is  the  pole  of  the  arc  AB. 

Let  0  be  the  centre  of  the  sphere,  and  draw  OA,  OB,  and  OP. 


194 


SOLID   GEOMETRY. 


[BK.  VIII. 


Then  since  PA  and  PB  are  quadrants,  the  angles  POA  and 
POB  are  right  angles. 

Therefore  PO  (by  337)  is  perpendicular  to  the  plane  AOB\ 
hence  P  is  the  pole  of  the  arc  AB.  Q.E.D. 


488,  COR.    The  polar  distance  of  a  great  circle  is  a  quadrant. 

489,  SCHOLIUM.     The  term  quadrant  in  Spherical  Geometry 
usually  signifies  a  quadrant  of  a  great  circle. 


SPHERICAL  ANGLES  AND  POLYGONS. 
DEFINITIONS. 

490.  The  Angle  between  two  intersecting  arcs  of  circles  on 
the  surface  of  a  sphere  is  the  diedral  angle  between  the  planes 
of  these  circles. 

A  Spherical  Angle  is  the  angle  between  two  intersecting  arcs 
of  great  circles  on  the  surface  of  a  sphere. 

491,  A  Spherical  Polygon  is  a  portion  of  the  surface  of  a 
sphere  bounded  by  three  or  more  arcs  of  great  circles. 

The  bounding  arcs  are  the  Sides  of  the  polygon ;  the  points 
of  intersection  of  the  sides  are  the  Vertices  of  the  polygon,  and 
the  angles  which  the  sides  make  with  each  other  are  the  Angles 
of  the  polygon. 

A  Diagonal  of  a  spherical  polygon  is  an  arc  of  a  great  circle 
joining  any  two  vertices  which  are  not  consecutive. 


§497.]         SPHERICAL  ANGLES  AND  POLYGONS.  195 

492.  A  Spherical  Triangle  is  a  spherical  polygon  of  three 
sides. 

A  spherical  triangle  is  Right  or  Oblique,  Scalene,  Isosceles, 
or  Equilateral,  in  the  same  cases  as  a  plane  triangle. 

493.  A  Spherical  Pyramid  is  a  portion  of  the  sphere  bounded 
by  a  spherical  polygon  and  the  planes  of  the  sides  of  the  poly- 
gon.    The  centre  of  the  sphere  is  the  Vertex  of  the  pyramid, 
and  the  spherical  polygon  is  its  Base, 

494.  Since  the  sides  of  a  spherical  polygon  are  arcs,  they 
are  usually  expressed  in  Degrees,  Minutes,  and  Seconds. 

495.  Two  spherical  polygons  are  Equal  if  they  can  be  applied 
one  to  the  other  so  as  to  coincide. 

496.  Two  spherical  polygons  are  Symmetrical  when  the  sides 
and  angles  of  the  one  are  respectively  equal  to  the  sides  and 
angles  of  the  other,  but  taken  in  the  reverse  order. 

Thus  the  spherical  triangles  ABC 
and  A'B'C'  are  symmetrical  if  the 
sides  AB,  BC,  and  CA  are  equal  to  /^C  C'^ 

AB,    B'O,   and    C'A',    respectively,    B-  ^B' 

and  the  angles  A,  B,  and  C  to  the  angles  A,  B',  and  C'. 

497.  Two  spherical  triangles  on  the  same  sphere  or  on  equal 
spheres  are  equal  (or  symmetrical),  under  the  same  conditions 
as  plane  triangles,  viz. : 

a.  When  they  have  two  sides  and  the  included  angle  equal. 

b.  When  they  have  two  angles  and  the  included  side  equal. 

c.  When  they  have  three  sides  equal. 


196 


SOLID   GEOMETRY. 


[BK.  VIII. 


If  the  equal  parts  are  in  the  same  order  as  in  ABC  and 
DEF,  equality  is  shown  by  superposition  as  in  Plane  Geometry. 

If  the  parts  are  in  the  reverse  order  as  ABC  and  D'E'F', 
construct  a  triangle,  DEF,  symmetrical  to  D'E'F1,  and  then  it 
can  be  shown  that  ABC  and  DEF  are  equal  by  superposition. 

PROPOSITION  II.     THEOREM. 

498.  A  spherical  angle  is  measured  by  the  arc  of  a  great 
circle  described  with  its  vertex  as  a  pole,  included  between  its 
sides  produced  if  necessary. 


Let  ABC  and  AB'C  be  two  intersecting  arcs  of  great  circles 
on  the  sphere  AC,  and  0  the  centre  of  the  sphere. 

Pass  the  plane  OBB'  perpendicular  to  AC  at  0,  intersecting 
the  planes  ABC  and  AB'C  in  the  radii  OB  and  OB',  and  the 
sphere  in  the  great  circle  BB'. 

To  prove  that  the  spherical  angle  BAB'  is  measured  by  the 
arc  BB'. 

Since  (by  359)  BOB'  is  a  plane  angle,  it  is  (by  368)  the 
measure  of  the  diedral  angle  BACB'. 

But  (by  174)  the  arc  BB'  is  the  measure  of  Z  BOB'. 

Therefore  the  spherical  angle  BAB'  is  measured  by  the 
arc  BB'.  Q.E.D. 

499.  COR.  1.  A  spherical  angle  is  equal  to  the  diedral  angle 
between  the  planes  of  the  two  circles. 


§  507.]         SPHERICAL  ANGLES  AND  POLYGONS.  197 

500.  COB.  2.     If  two  arcs  of  great  circles  cut  each  other,  their 
vertical  angles  are  equal. 

501.  COR.  3.      The  angles  of  a  spherical  polygon  are  equal  to- 
the  diedral  angles  between  the  planes  of  the  sides  of  the  polygon. 

502.  Because  the  planes  of  all  great  circles  pass  through 
the  centre  of  the  sphere,  therefore  the  planes  of  the  sides  of  a 
spherical  polygon  form  a  polyedral  angle 

at  the  centre  0  whose  face-angles  AOB, 
BOC,  etc.,  are  measured  by  the  sides  AB, 
BC,  etc.,  of  the  polygon,  and  whose  diedral 
angles  OA,  OB,  etc.,  are  equal  to  the  angles 
A,  B,  etc.,  of  the  spherical  polygon  ABC, 
etc. 

We  may  therefore  speak  of  all  the  parts 
of  a  spherical  polygon  as  Angles,  meaning 
thereby  the  face-angles,  and  the  diedral  angles  between  the 
faces,  of  the  polyedral  angle  whose  vertex  is  the  centre  of  the 
sphere,  and  whose  base  is  the  spherical  polygon. 

503.  SCHOLIUM.     Since  the  sides  and  angles  of  a  spherical 
polygon  are  measured  by  the  face  and  diedral  angles  of  the 
polyedral   angle  corresponding  to  the  polygon,  ice  may,  from 
any  property  of  polyedral  angles,  infer  an  analogous  property  of 
spherical  polygons. 

504.  Each  side  of  a  spherical  triangle  is  less  than  the  sum  of 
the  other  two.  (384) 

505.  Any  side  of  a  spherical  polygon  is  less  than  the  sum  of 
the  other  sides. 

506.  Tlie  sum  of  the  sides  of  a  spherical  polygon  is  less  than 
§60°.  (385) 

507.  Two   mutually   equilateral    triangles    on    equal    spheres 
are  mutually  equiangular,  and   are  equal,  or  symmetrical  and 
equivalent. 


198 


SOLID   GEOMETRY 


[Biv.  VIII. 


508.    In  an  isosceles  spherical  triangle,  the  angles  opposite  the 
equal  sides  are  equal. 


509.  The  arc  of  a  great  circle  drawn  from  the  vertex  of  an 
isosceles  spherical  triangle  to  the  middle  of  the  base  is  perpendicu- 
lar to  the  base,  and  bisects  the  vertical  angle. 


510.  If  with  the  vertices  of  a  spheri- 
cal triangle  as  poles  arcs  of  great  cir- 
cles are  described,  a  spherical  triangle 
is  formed  which  is  called  the  Polar 
Triangle  of  the  first. 

Thus,  if  A,  B,  and  C  are  the  poles 
of  the  arcs  B'C',  C'A',  and  A'B',  then 
A'B'C'  is  the  polar  triangle  of  ABC. 


PROPOSITION  III.     THEOREM. 

511.    If  the  first  of  two  spherical  triangles  is  the  polar  triangle 
of  the  second,  then  the  second  is  the  polar  triangle  of  the  first. 

Let  A'B'C'  be   the   polar   triangle   of 
ABC. 

To  prove  that  ABC  is  the  polar  tri- 
angle of  A1  B'C'. 

Since  B  is  the  pole  of  the  arc  A'C',  it    n 
is  a  quadrant's  distance  from  A'.     Also, 
since  C  is  the  pole  of  the  arc  A'B',  it  is  a  quadrant's  distance 
from  A'. 

Therefore  A  is  a  quadrant's  distance  from  B  and  (7,  hence 
from  the  arc  BC,  or  is  the  pole  of  the  arc  BC. 

Similarly,  B'  can  be  shown  to  be  the  pole  of  the  arc  AC,  and 
C'  the  pole  of  AB. 

Hence  ABC  is  the  polar  triangle  of  A'B'C'.  Q.E.D. 


§  514.]          SPHERICAL  ANGLES  AND  POLYGONS.  199 

PROPOSITION  IV.     THEOREM. 

512.  In  two  polar  triangles,  each  angle  of  one  is  the  supple- 
ment of  the  side  opposite  to  it  in  the  other. 

Let  ABC  and  A'B'C'  be  a  pair  of  polar  triangles  in  which 
A,  B,  C,  A',  B'}  and  C'  are  the  angles,  and  a,  b,  c,  a',  I)',  and 
c'  the  sides. 

To  prove  that 

A  =  180°  -  a',  A'  =  180°  -  a, 
£=180°  -6',  B'  =  180°  -b, 
C  =  180°  -  c',  C"  =  180°  -  c. 

Produce  the  arc  AB  until  it  meets 
B'C'  in  J9,  and  .4(7  until  it  meet  B'C' 
in  ,£. 

Since  Z»'  is  the  pole  of  AC,  it  will  be  a  quadrant's  distance 
from  E,  or  B'E  =  90°  ;  likewise,  OD  =  90°. 

Hence  B'E  +  C'D  =  180°, 

or  £'£>  +  DE+  OD  =  180°  ; 

that  is,  B'C'  +  DE  =  180°. 

But  (by  49$)  DE  is  the  measure  of  Z.  A, 
therefore  ZA  +  a'  =  180°, 

or  Z.A  =  180°  -  a'. 

The  other  relations  may  be  proved  in  a  similar  manner.  Q.E.D. 

513.  SCHOLIUM.    Two  spherical  polygons  are  mutually  equi- 
lateral or  mutually  equiangular  when  the  sides  or  angles  of 
one  are  equal  respectively  to  the  sides  or  angles  of  the  other, 
whether  taken  in  the  same  or  in  the  reverse  order. 


514.    COR.    If  two  splierii'dl  triangles  are  mutually  equiangular, 
their  polar  triangles  are  mutually  equilateral. 


200 


SOLID   GEOMETRY. 


[BK.  VIII. 


Since  (by  512),  any  two  homologous  sides  in  the  polar 
triangles  are  supplements  of  equal  angles  in  the  original  tri- 
angles, hence  they  are  equal. 

515.  If  two  spherical  triangles  are  mutually  equilateral, 
their  polar  triangles  are  mutually  equiangular. 


PROPOSITION  V.     THEOREM. 

516.  The  sum  of  the  angles  of  a  spherical  triangle  is  greater 
than  two,  and  less  than  six,  right  angles. 

Let  ABC  be  any  spherical  triangle. 
To  prove  that 

A  +  B  +  C  >  180°  <  540°. 

Let  A'B'C'  be   the   polar   triangle, 
then  (by  512) 

^L-f-a'=180°,  B+b'=180°,  C+cf  =  180°, 
or  A  +  B  +  C  +  a!+V  + 

But  (by  506) 

a'  +  6'  +  c'  <  360°,  and  a'  +  b'  +  c'  >  0. 

Therefore,  by  subtraction, 

A  +  B  +  C  >  180°  <  540°.  Q.E.D. 

517.  SCHOLIUM.     The  amount  by  which  the  three  angles  of 
a  spherical  triangle  exceeds  180°  is  called  the  Spherical  Excess. 

518.  COR.     A  spherical  triangle  may  have  two,  or  even  three, 
right  angles;  also  two,  or  even  three,  obtuse  angles. 

519.  If  a  spherical  triangle  has  two  right  angles,  it  is  called 
a  Bi-rectangular  Triangle;  and  if  a  spherical  triangle  has  three 
right  angles,  it  is  called  a  Tri-rectangular  Triangle.     Its  surface 
is  one-eighth  part  of  the  surface  of  the  sphere. 


§  523.] 


SPHERICAL  ANGLES  AND  POLYGONS. 


201 


520.  A  Lune  is  a  portion  of  the  surface 
of  a  sphere  included  between  two  semi- 
circumferences  of  great  circles;  as  ACBD. 

The  Angle  of  a  lime  is  the  angle  between 
the     semi-circumferences    which     form     its 
sides;     as    the    angle    CAD,    or    the    angle 
COD. 


521.  On   the  same,  or  on   equal,   spheres,   lunes  of   equal 
angles  are  equal,  as  they  are  evidently  superposable. 

522.  A  Spherical  Wedge,  or  Ungula,  is  the  part  of  a  sphere 
bounded  by  a  lime  and  the  planes  of  its  sides ;  as  AOBCD. 

The  diameter  AB  is  called  the  Edge  of  the  ungula,  and  the 
lime  ACBD  is  called  its  Base. 


PROPOSITION  VI.     THEOREM. 

523.    The  area  of  a  lune  is  to  the  surface  of  the  sphere  as  the 
angle  of  the  lune  is  to  four  right  angles. 

Let  ACBD  be  a  lune,  and  ECD H  the 
great  circle  whose  poles  are  A  and  B; 
let  L  be  the  area  of  the  lune,  and  5  the 
surface  of  the  sphere,  and  A  the  angle 
CAD,  or  the  angle  of  the  lune. 

To  prove  that 


L 

S 


L      arc  CD 


360 


0> 


or  —  = 


S      ECDH 


since  angle  A  is  measured  by  arc  CD,  and  ECDH  is  the  cir- 
cumference. Apply  a  common  measure  to  CD  and  ECDH, 
and  suppose  it  is  contained  n  times  in  CD  and  m  times  in 

ECDH,  then      CD     =--. 
ECDH      m 

Through  these  points  of  division   pass  great  circles ;  then 


202  SOLID  GEOMETRY.  [BK.  VIII. 

the  lime  ACBD  will  contain  n  equal  limes,  and  the  entire 
sphere  m  equal  lunes, 

L      n 

or  —  —  — 

S     m 

rpi       f  L      arc  CD        A 

Inereiore  -  = = Q.E.D. 

S     ECDH     360° 

The  student  can  supply  the  proof  for  the  case  when  CD  and 
ECDH  are  incommensurable. 


524.  COR.  Let  A  denote  the  numerical  measure  of  the 
angle  of  a  lune  referred  to  a  right  angle  as  the  unit,  and  T  the 
area  of  the  tri-rectangular  triangle. 

Then  since  the  surface  of  the  sphere  is  expressed  by  8  T,  we 
have  (by  523) 

=  -    or  L  =  2Ax  T. 


—     -, 

ST     4 

That  is,  if  the  unit  of  measurement  for  angles  is  the  right 
angle,  the  area  of  a  lune  is  equal  to  twice  its  angle  multiplied 
by  the  area  of  the  tri-rectangular  triangle. 

For  example,  if  A  =  60°  —  -|  of  a  right  angle,  its  area  would 
be  J  of  the  area  of  the  tri-rectangular  triangle.  Then  if  the 
surface  of  the  sphere  were  120  square  inches,  the  area  of  the 
tri-rectangular  triangle  is  15  square  inches,  |-  of  which  is  20 
square  inches  or  the  area  of  the  lune. 


EXERCISES. 

1.  Show  that  in  a  spherical  triangle,  each  side  is  greater  than  the  dif- 
ference between  the  other  two. 

2.  If  the  radius  of  the  sphere  is  12,  what  is  the  linear  length  of  the 
sides  of  the  triangle  whose  angular  measures  are  40°,  60°,  and  80°  ? 

3.  Find  the  area  of  a  lune  when  the  angle  is  135°,  and  the  surface  of 
the  sphere  300  square  inches. 


§  526.]  THE   SPHERE.  203 


PROPOSITION  VII.     THEOREM. 

525.  If  two  arcs  of  great  circles  BAB1  and  CAC'  intersect 
each  other  on  the  surface  of  a  hemisphere,  the  sum  of  the  oppo- 
site triangles  ABC  and  AB'C'  is  equivalent  to  a  lune  ichose  angle 
is  equal  to  the  angle  BAG  included  between  the  given  arcs. 


Draw  the  diameters  AA,  BB\  CO. 

Since  ABA  is  a  semicircle,  it  is  equal  to  BAB1 ;  sub- 
tract the  portion  BA,  and  we  have  arc  A'B  =  AB' ;  likewise 
arc  A'C  =  AC'j  and  BC  =  B'C')  both  being  measures  of  the 
equal  vertical  angles. 

Therefore  A'BC  =  AB'C'. 

Adding  BAG,  we  have 

A'BC  +  BAG  =  BAG  +  AB'C', 
or  lune  ABAC  =  BAG  +  AB'C'.  Q.E.D. 

PROPOSITION  VIII.     THEOREM. 

526.  The  area  of  a  spherical  triangle  is  equal  to  its  spherical 
excess. 

Let  A,  B,  C  be  the  numerical  meas- 
ures   of    the    angles    of    the    spherical       / 
triangle  ABC-,    let  the  right  angle  be      ,{/ 
the  unit  of  angular  measure,  and  the 
tri-rectangular   triangle   T  be  the  unit 
of  areas. 


204  SOLID   GEOMETRY.  [BK.  VIII. 

To  prove  that 

Area  ABC  =  (A  +  B+C-2)xT. 

Continue  any  side,  say  AB,  so  as  to  complete  the  great 
circle,  and  produce  the  other  sides  until  they  meet  this  circle 
in  B'  and  A'. 

Area  ABC  +  ABC  =  lune  ABAC  =  2  A  x  T- 
likewise  ABC  +  AB'G  =  lune  ABCB'  =  2B  x  T, 
and  ABC+A'B'C  =  lune  ACBC'  =2C  x  T. 

By  addition, 

3 ABC  +  A'BC  +  AB'C+AB'C  =  (2A  +  2B  +  2C)  x  T. 
But  ABC  +  A'BC  +  AB'C  +  A'B'C=  the  hemisphere  =  4  T. 
Therefore     2ABC+4  T  =  (2A  +  2B  +  2C)  x  T, 

ABC+2T=(A  +  B+C)  x  T, 
or  ABC  =  (A  +  B  +C-  2)  x  T.  Q.E.D. 

The  greater  the  excess  of  A-\-  B  +  C  over  2  right  angles, 
the  greater  will  be  the  area. 

527.  COR.      The  area  of  a  spherical  polygon  is  equal  to  its 
spherical  excess. 

THE   SPHERE. 

528.  A  Zone  is  a  portion  of  the  surface  of  a  sphere  included 
between  parallel  planes. 

The  circumference  of  the  circles  which  bound  the  zone  are 
called  the  Bases,  and  the  distance  between  their  planes  the 
Altitude. 

One  of  the  bases  may  be  a  tangent  plane. 

529.  A  Spherical  Segment  is  a  portion  of  the  volume  of  the 
sphere  included  between  two  parallel  planes ;  the  planes  are 
the  Bases,  and  their  distance  apart  is  the  Altitude. 

529  a.  A  Spherical  Sector  is  the  portion  of  a  sphere  gener- 
ated by  the  revolution  of  a  circular  sector  about  the  diameter 
of  its  circle  as  an  axis. 


§531.] 


THE  SPHERE. 


205 


c 


530.  Let  the  sphere  be  generated  by  the  revolution  of  the 
semicircle  AGDEFB  about  its  diameter  AB  as  an  axis ;  and 
let  CG  and  DH  be  drawn  perpendicular 
to  the  axis.  The  arc  CD  generates  a  zone 
whose  altitude  is  GH,  and  the  figure 
CDIIG  generates  a  spherical  segment 
whose  altitude  is  GH.  The  circumfer- 
ences generated  by  the  points  C  and  D 
are  the  bases  of  the  zone,  and  the  circles 
generated  by  CG  and  DH  are  the  bases 


of  the  segment. 


PROPOSITION  IX.     THEOREM. 

531.  The  area  generated  by  the  revolution  of  a  straight  line 
about  an  axis  in  its  plane  is  equal  to  the  projection  of  the  line 
on  the  axis,  multiplied  by  the  circumference  of  a  circle  ivhose 
radius  is  the  length  of  the  perpendicular  erected  at  the  middle 
point  of  the  line  and  terminating  in  the  axis. 

M 
C 

H 

D 
F 

Let  AB  be  the  straight  line  revolving  about  the  axis  FM. 
CD  its  projection  on  FM,  and  EF  the  perpendicular  erected 
at  the  middle  point  of  AB,  terminating  in  the  axis. 

To  prove  that 

area  generated  by  AB  =  CD  x  2  -n-  x  EF. 

Draw  AG  parallel  to  CD,  and  EH  perpendicular  to  CD. 
The  area  generated  by  AB  is  the  lateral  surface  of  a  frustum 


206 


SOLID   GEOMETR  Y. 


[BK.  VIII. 


of  a  cone  of  revolution,  with  AC  and  BD  as  radii  of  the  upper 
and  lower  bases. 
Therefore  (by  461) 

area  AB  =  AB  x  2  -n-  x  EH. 
The  triangles  are  (by  64  and  218)  similar,  hence 

or 


A(JT 


x  EH=AG  x  ^=  CD  X  EF. 


Substituting  this  value  for  AB  x  EH,  we  have 

area  AB  =  CD  x  2  -n-  x  EF.  Q.E.D 

PROPOSITION  X.     THEOREM. 

532.    The  area  of  the  surface  of  a  sphere  is  equal  to  the  product 
of  its  diameter  by  the  circumference  of  a  great  circle. 

Let  the  sphere  be  generated  by  the  revolu- 
tion of  the  semicircle  ABDF  about  the  diame- 
ter AF,  let  0  be  the  centre,  R  the  radius,  and 
denote  the  surface  of  the  sphere  by  S. 

To  prove        S  =  AF  x  2  irR. 

Inscribe  in  the  semicircle  a  regular  semi- 
polygon  ABCDEF  of  any  number  of  sides. 

Draw  Bb,  Cc,  Dd,  etc.,  perpendicular  to  AF, 
and  OH  perpendicular  to  BA ;  then  (from  146) 
AB  is  bisected  in  H. 

Then  (from  531)     area  AB  =  Ab  x  2  TT  x  OH. 
Likewise  area  BC  =  be  x  2  -n-  x  OG-,  etc. 

But  (by  148)  OG  =  OH. 

Therefore     area  generated  by  ABC =  Ac  x  2  TT  x  OH. 

Now  the  sum  of  the  projections  of  the  sides  of  the  semi- 
polygon  make  up  the  diameter  AF,  hence  the 

area  generated  by  ABCDEF  =  ^LF  x  2  TT  x  OH. 


§  536.]  THE   SPHERE.  207 

Now  let  the  number  of  sides  of  the  inscribed  semi-polygon 
be  indefinitely  increased. 

The  semi-perimeter  will  approach  the  semi-circumference  as 
its  limit,  and  OH  will  approach  the  radius  R  as  its  limit. 

Therefore  the  surface  of  revolution  will  approach  the  surface 
of  the  sphere  as  its  limit ;  hence 

£  —  AF  x  2  TT  x  R.  Q.E.D. 

533.  COB.  1.     Since       AF=2R, 

3  =  2  R  x  2  irR  =  4  TrR\ 

Therefore,  the  area  of  the  surface  of  a  spliere,  is  equal  to  the 
area  of  four  great  circles  of  that  sphere. 

534.  COR.  2.     The  areas  of  the  surfaces  of  two  spheres  are  to 
each  other  as  the  squares  of  their  radii,  or  as  the  squares  of  their 
diameters. 

535.  The  area  of  a  zone  is  equal  to  the  product  of  its  altitude 
by  the  circumference  of  a  great  circle. 


PROPOSITION  XI.     THEOREM. 

536.  The  volume  of  a  sphere  is  equal  to  the  area  of  its  surface 
multiplied  by  one-third  of  its  radius. 

Let  V  denote  the  volume  of  a  sphere,  S  the  area  of  its  sur- 
face, and  R  its  radius. 

To  prove  that  V=  S  x  -J-  R. 

Conceive  any  polyedron  as  circumscribed  about  the  sphere ; 
then  if  the  number  of  faces  be  indefinitely  increased,  each  face 
will  be  diminished,  and  the  limit  of  the  surface  of  the 
polyedron  is  the  surface  of  the  sphere,  and  the  limit  of  the 
volume  of  the  polyedron  is  the  volume  of  the  sphere. 

Let  each  vertex  of  the  polyedron  be  joined  to  the  centre  of 


208  SOLID   GEOMETEY.  [BK.  VIII. 

the  sphere,  then  the  entire  polyedron  will  be  made  up  of  pyra- 
mids. 

The  volume  of  the  entire  polyedron  is  equal  to  the  sum  of 
the  volumes  of  the  pyramids  ;  that  is,  the  sum  of  the  bases 
multiplied  by  one-third  of  the  common  altitude,  or  the  radius  of 
the  sphere. 

Since  this  is  true  whatever  the  number  of  faces  may  be,  the 
limiting  volume  will  be  equal  to  the  limiting  surface  multiplied 
by  one-third  of  the  radius,  or 

V=  S  x  I  R. 

537.   COR.  1.     Since  (by  533) 

S  =  4 


and  V= 

538.  COR.  2.     The  volumes  of  two  spheres  are  to  each  other  as 
the  cubes  of  their  radii. 

539.  COR.  3.     The  volume  of  a  spherical  sector  is  equal  to  the 
area  of  the  zone  which  forms  its  base  multiplied  by  one-third  the 
radius  of  the  sphere. 

For  a  spherical  sector,  like  the  entire  sphere,  may  be  con- 
ceived as  consisting  of  an  indefinitely  great  number  of  pyra- 
mids whose  bases  make  up  its  base,  and  whose  common  altitude 
is  the  radius  of  the  sphere. 

540.  COR.  4.     The  volume  of  the  cylinder  circumscribed 
about  a  sphere  =  2  rrR3. 

Therefore,  the  volume  of  a  sphere  is  equal  to  two-thirds  the 
volume  of  the  circumscribing  cylinder. 

EXERCISES. 

1.  Find  the  surface  and  volume  of  a  sphere  whose  radius  is  12. 

2.  Find  the  diameter  and  surface  of  a  sphere  whose  volume  is  896. 


§  541.]  FORMULA.  209 

3.  If  the  radii  of  the  bases  of  a  spherical  segment  are  4  and  6  and  the 
altitude  is  5,  find  the  radius  of  the  sphere. 

4.  Find  the  number  of  cubic  feet  in  a  log  18  feet  long  and  6J  feet  in 
diameter. 

5.  Find  the  number  of  gallons  in  a  cistern  6  feet  in  diameter  and 
10  feet  deep,  if  231  cu.  in.  make  a  gallon. 

6.  Find  the  weight  of  an  iron  shell  4  inches  in  diameter,  the  iron 
being  1^  in.  thick,  and  weighing  \  of  a  pound  to  the  cubic  inch. 

7.  Show  that  the  surface  of  a  sphere  is  equal  to  the  lateral  surface  of 
its  circumscribing  cylinder. 

541.  FORMULAE. 

NOTATION. 

S  =  surface  (or  area).  r  =  radius  of  inscribed  circle. 

V  =  volume.  R  =  radius  of  circle  (general). 

h  =  altitude.  R'  =  radius  of  upper  base. 

b  —  lower  base  (linear).  D  =  diameter. 

b'  —  upper  base  (linear).  L  =  slant  height. 

s  =  i  (a  +  b  +  c).  C  =  circumference. 

P=  perimeter.  B  =  area  of  base. 
P'  =  perimeter  of  upper  base.     B'  =  area  of  upper  base. 

Parallelogram ;  S  =  h  X  b.  (251) 

Triangle ;  S  =  \  h  x  b.  (254) 

(271) 


Trapezoid ;                                S  =  ^  h  [5  4-  6'].  (258) 

Polygon;                                  8  =  ±Pxr.  (292) 

PiTplp  •                                                C1 9^-vP--vD  f30T\ 

V^llOlC  >                                                                           \_/  +J  7T   /s.    JLv  7T    /\    jL/.  Ic/V/t-M 

S  =  7rxR*.  (305) 

Sector ;                                      S  =  1  arc  x  R.  (306) 


210  SOLID   GEOMETRY.  [BK.  VIII. ,  §  541. 

TT  =  3.141592635...  (310) 
log  TT  =  0.4971498726  ... 

Eight  Prism ;  lateral  S  =  P  X  h.  (416) 

.      V=  B  x  h.  (428) 

Cylinder ;  lateral  S  =  2  TT  x  R  x  h.  (417) 

V=  TT  x  R2  x  h.  (433) 

Parallelepiped ;  F=  B  x  h.  (425) 

Pyramid ;  lateral  S  =  ±  P  x  L.  (458) 

V=  i  B  x  h.  (465) 

Cone ;  lateral  S  =  TT  x  R  x  L.  (460) 

F=  i  TT  x  ,R2  x  7i.  (466) 

Frustum  of  a  pyramid ;  lateral  S  =  ±(P+  P')  x  L.  (459) 


F=J-ft[-B+-B'  +  V.Bx.B'].  (469) 
Frustum  of  a  cone ;  lateral  /S  =  TT  (R-\-Rr)  x  L.  (461) 

F=  -|-7rX/i[^2H-^'2  +  ^tXP'].  (471) 
Sphere;  >S  =  47rX^2.  (533) 

F-^TrxD3.  (537) 

Zone ;  8  =  2  TT  x  R  x  h.  (535) 


APPENDIX. 

ADDITIONAL   EXERCISES   ON   BOOK   I. 

1.  Each  exterior  angle  of  an  equilateral  triangle  equals  how  many 
times  each  interior  angle  ? 

2.  From  a  point  without  a  line,  show  that  only  two  oblique  lines  can 
be  drawn  so  as  to  make  equal  angles  with  the  given  line. 

3.  If  a  straight  line  meets  two  parallel  straight  lines,  and  the  two 
interior  angles  on  the  same  side  are  bisected,  show  that  the  bisectors 
meet  at  right  angles. 

4.  How  many  sides  has  a  polygon,  the  sum  of  whose  interior  angles 
is  equal  to  the  sum  of  its  exterior  angles  ? 

5.  The  sum  of  the  three  medial  lines  of  a  triangle  is  less  .than  the 
perimeter,  and  greater  than  half  the  perimeter  of  the  triangle. 

6.  How  many  sides  has  a  polygon,  the  sum  of  whose  interior  angles 
is  double  that  of  its  exterior  angles  ? 

7.  If  BC,  the  base  of  an  isosceles  triangle  ABC,  is  produced  to  any 
point  Z),  show  that  AD  is  greater  than  either  of  the  equal  sides. 

8.  Prove  that  any  point  not  in  the  bisector  of  an  angle  is  unequally 
distant  from  its  sides. 

9.  The  diagonals  of  a  rhombus  are  perpendicular  to  each  other  and 
bisect  the  angles  of  the  rhombus. 

10.  The  angles  a,  &,  c,  d  are  such  that  a-\-b  +  c  +  d  =  &  straight  angle, 
and  a  =  26  =  4c  =  8d.     How  many  degrees  in  a,  &,  c,  d ? 

11.  If  an  angle  at  the  base  of  an  isosceles  triangle  is  n  times  the  verti- 
cal angle,  what  fraction  is  the  latter  of  a  straight  angle  ? 

12.  The  straight  line  A  E  which  bisects  the  angle  exterior  to  the  vertical 
angle  of  an  isosceles  triangle  ABC,  is  parallel  to  the  base  BC. 

13.  The  lines  joining  the  middle  points  of  the  sides  of  a  triangle  divide 
the  triangle  into  four  equal  triangles. 

211 


212  PLANE  GEOMETRY. 

14.  If  both  diagonals  of  a  parallelogram  are  drawn,  of  the  four  tri- 
angles thus  formed  those  opposite  are  equal. 

15.  In  a  triangle  ABC,  if  AC  is  not  greater  than  AS,  show  that  any 
straight  line  drawn  through  the  vertex  A  and  terminated  by  the  base  BC 
is  less  than  AB. 

16.  The  lines  joining  the  middle  points  of  the  sides  of  a  rhombus, 
taken  in  order,  include  a  rectangle. 

17.  The  lines  joining  the  middle  points  of  the  sides  of  any  quadrilateral, 
taken  in  order,  enclose  a  parallelogram. 

18.  In  any  right  triangle,  the  straight  line  drawn  from  the  vertex  to 
the  middle  point  of  the  hypotenuse  is  equal  to  half  the  hypotenuse. 

19.  If  a  diagonal  of  a  quadrilateral  bisects  two  angles,  the  quadrilateral 
has  two  pairs  of  equal  sides. 

20.  If  one  of  the  acute  angles  of  a  right  triangle  is  double  the  other, 
the  hypotenuse  is  double  the  shorter  side. 

21.  The  perimeter  of  a  quadrilateral  is  less  than  twice  the  sum  of  its 
two  diagonals. 

22.  If,  in  a  quadrilateral,  two  opposite  sides  are  equal,  and  the  two 
angles  which  a  third  side  makes  with  the  equal  sides  are  equal,  then  the 
other  two  angles  are  equal  also. 

23.  The  straight  lines  joining  the  middle  points  of  the  opposite  sides 
of  any  quadrilateral  bisect  each  other. 

24.  Any  two  sides  of  a  triangle  are  together  greater  than  twice  the 
straight  line  drawn  from  the  vertex  to  the  middle  point  of  the  third  side. 

25.  If  ABC  is  an  equilateral  triangle,  and  if  BD  and  CD  bisect  the 
angles  B  and  C,  the  lines  DE,  DF,  parallel  to  AB,  AC,  respectively, 
divide  BC  into  three  equal  parts. 

26.  If  from  a  variable  point  in  the  base  of  an  isosceles  triangle  par- 
allels to  the  sides  are  drawn,  a  parallelogram  is  formed  whose  perimeter 
is  constant. 

27.  The  diagonals  of  a  square  or  rhombus  bisect  each  other  at  right 
angles,  and  bisect  the  angles  whose  vertices  they  join. 

28.  If  BE  bisects  the  angle  B  of  a  triangle  ABC,  and  CE  bisects  the 
exterior  angle  ACD,  the  angle  E  is  equal  to  one-half  the  angle  A. 


APPENDIX.  213 

29.  The  sum  of  the  perpendiculars  dropped  from  any  point  within  an 
equilateral  triangle  to  any  one  of  the  three  sides  is  constant,  and  equal 
to  the  altitude. 

30.  The  median  to  any  side  of  a  triangle  is  less  than  the  half-sum  of 
the  other  two  sides,  but  greater  than  half  of  the  difference  between  their 
sum  and  the  third  side. 

31.  If  the  bisectors  of  two  angles  of  an  equilateral  triangle  meet,  and 
from  the  point  of  meeting  lines  be  drawn  parallel  to  any  two  sides,  these 

lines  will  trisect  the  third  side. 
/ 

32.  The  sum  of  four  lines  drawn  to  the  vertices  of  a  quadrilateral  from 
any  point  except  the  intersection  of  the  diagonals,  is  greater  than  the 
sum  of  the  diagonals. 

33.  In  a  quadrilateral,  the  sum  of  either  pair  of  opposite  sides  is  less 
than  the  sum  of  its  two  diagonals. 

34.  The  interior  angle  of  a  regular  polygon  is  five-thirds  of  a  right 
angle.     Find  the  number  of  sides  in  the  polygon. 

35.  The  exterior  angle  of  a  regular  polygon  is  one-fifth  of  a  right 
angle.     Find  the  number  of  sides  in  the  polygon. 

36.  If  one  side  of  a  regular  hexagon  is  produced,  show  that  the  exterior 
angle  is  equal  to  the  angle  of  an  equilateral  triangle. 

37.  How  many  braces  would  it  take   to   stiffen  a  three-sided  plane 
figure  ?    Four-sided  ?     Five-sided  ? 

38.  If  from  any  point  equidistant  from  two  parallels  two  transversals 
are  drawn,  they  will  cut  off  equal  segments  of  the  parallels. 

39.  If  a  quadrilateral  have  two  of  its  opposite  sides  parallel,  and  the 
other  two  equal   but  not  parallel,  any  two  of   its  opposite  angles  are 
together  equal  to  two  right  angles. 

40.  The  sum  of  the  perpendiculars  from  any  point  in  the  interior  of 
an  equilateral  triangle  is  equal  to  the  distance  of  any  vertex  from  the 
opposite  side. 

41.  A  line  is  drawn   terminated  by  two  parallel  lines ;    through  its 
middle  point  any  line  is  drawn  and  terminated  by  the  parallel  lines. 
Show  that  the  second  line  is  bisected  at  the  middle  point  of  the  first. 


214  PLANE  GEOMETRY. 


ADDITIONAL  EXERCISES   ON   BOOK  II. 

1.  If  an  isosceles  triangle  be  constructed  on  any  chord  of  a  circle,  its 
vertex  will  be  in  a  diameter,  or  a  diameter  produced. 

2.  The  perimeter  of  an  inscribed  equilateral  triangle  is  equal  to  half 
the  perimeter  of  the  circumscribed  equilateral  triangle. 

3.  If  two  equal  chords  intersect,  their  segments  are  severally  equal. 

4.  The  perpendiculars  from  the  angles  upon  the  opposite  sides  of  the 
triangle  are  the  bisectors  of  the  angles  of  the  triangle  formed  by  joining 
the  feet  of  the  perpendiculars. 

5.  A  straight  line  will  cut  a  circle,  or  lie  entirely  without  it,  according 
as  its  distance  from  the  centre  is  less  than,  or  greater  than,  the  radius  of 
the  circle. 

6.  The  bisectors  of  the  angle  contained  by  the  opposite  sides  (pro- 
duced) of  an  inscribed  quadrilateral,  intersect  at  right  angles. 

7.  A,  B,  and  C  are  three  points  on  the  circumference  of  a  circle,  the 
bisectors  of  the  angles  A,  B,  and  G  meet  at  Z>,  and  AD  produced  meets 
the  circle  in  E ;  prove  that  ED  =  EC. 

8.  If  a  variable  tangent  meets  two  parallel  tangents  it  subtends  a 
right  angle  at  the  centre. 

9.  If  through  any  point  in  a  radius  two  chords  are  drawn,  making 
equal  oblique  angles  with  it,  these  chords  are  equal. 

10.  Two  circles  are  tangent  internally  at  P,  and  a  chord  AB  of  the 
larger  circle  touches  the  smaller  at  (7.     Prove  that  PC  bisects  the  angle 
APB. 

11.  All  chords  of  a  circle  which  touch  an  interior  concentric  circle  are 
equal,  and  are  bisected  at  the  points  of  contact. 

12.  If  a  straight  line  cuts  two  concentric  circles,  the  parts  of  it  inter- 
cepted between  the  two  circumferences  are  equal. 


APPENDIX.  215 

13.  The  angle  formed  by  two  tangents  drawn  to  a  circle  from  the 
same  point,  is  supplementary  to  that  formed  by  the  radii  to  the  points  of 
contact. 

14.  In  two  concentric  circles  any  chord  of  the  outer  circle,   which 
touches  the  inner,  is  bisected  at  the  point  of  contact. 

15.  If  through  the  points  of  intersection  of  two  circumferences,  parallels 
be  drawn  to  meet  the  circumferences,  these  parallels  will  be  equal. 

16.  Through  one  of  the  points  of  intersection  of  two  circles  a  diameter 
of  each  circle  is  drawn.     Prove  that  the  straight  line  joining  the  ends  of 
the  diameters  passes  through  the  other  point  of  intersection. 

17.  If  a  circle  is  inscribed  in  a  trapezoid  that  has  equal  angles  at  the 
base,  each  nonparallel  side  is  equal  to  half  the  sum  of  the  parallel  sides. 

18.  If  two  circles  touch  externally  at  P,  the  straight  line  joining  the 
extremities   of  two  parallel  diameters,   towards   opposite  parts,    passes 
through  P. 

19.  OC  is  drawn  from  the  centre  0  of  a  circle  perpendicular  to  a  chord 
AB.     Prove  that  the  tangents  at  A,  B,  intersect  in  OC  produced. 

20.  Prove  that  two  of  the  straight  lines  which  join  the  ends  of  two 
equal  chords  are  parallel,  and  the  other  two  are  equal. 

21.  If  two  pairs  of  opposite  sides  of  a  hexagon  inscribed  in  a  circle  are 
parallel,  the  third  pair  of  opposite  sides  are  parallel. 

22.  To  construct  a  triangle  having  given  the  two  exterior  angles  and 
the  included  side. 

23.  To  divide  a  right  angle  into  three  equal  parts. 

24.  Construct  an  isosceles  triangle  having  its  sides  each  double  the 
length  of  the  base. 

25.  Through  a  given  point  to  draw  a  line  making  a  given  angle  with 
a  given  line. 

26.  Construct  a  right  triangle,  having  given  an  arm  and  the  altitude 
from  the  right  angle  upon  the  hypotenuse. 

27.  To  draw  a  line  through  a  given  point,  so  that  it  shall  form  with 
the  sides  of  a  given  angle  an  isosceles  triangle. 

28.  From  a  given  point  in  a  given  line  to  draw  a  line  making  an  angle 
supplemental  to  a  given  angle. 


216  PLANE  GEOMETET. 

29.  Through  a  given  point  P  within  a  given  angle  to  draw  a  straight 
line,  terminated  by  the  sides  of  the  angle,  which  shall  be  bisected  at  P. 

30.  Through  a  given  point  to  draw  a  line  which  shall  make  equal 
angles  with  the  two  sides  of  a  given  angle. 

31.  Construct  an  equilateral  triangle  having  a  given  altitude  AB. 

32.  To  construct  an  isosceles  triangle,  having  given  the  base  and  the 
opposite  angle. 

33.  On  a  given  straight  line  as  hypotenuse,  construct  a  right  triangle 
having  one  of  its  acute  angles  double  the  other. 

34.    The  bisectors  of  the  angles  of  a  circumscribed  quadrilateral  pass 
through  a  common  point. 

35.  Draw  a  straight  line  equally  distant  from  three  given  points. 

36.  Find  the  bisector  of  the  angle  that  would  be  formed  by  two  given 
lines,  without  producing  the  lines. 

37.  In  any  side  of  a  triangle,  find  the  point  which  is  equidistant  from 
the  other  two  sides. 

38.  Through  two  given  points  to  draw  the  two  lines  forming,  with  a 
given  line,  an  equilateral  triangle. 

39.  From  two  given  points  to  draw  lines  making  equal  angles  with 
a  given  line,  the  points  being  on  (1)  the  same  side  of  the  given  line, 
(2)  opposite  sides  of  the  given  line. 

40.  In  any  side  of  a  triangle,  find  the  point  from  which  the  lines  drawn 
parallel  to  the  other  two  sides  are  equal. 

41.  To  draw  a  tangent  to  a  given  circle  so  that  it  shall  be  parallel  to 
a  given  straight  line. 

42.  With  a  given  point  as  centre,  describe  a  circle  which  shall  be 
divided  by  a  given  straight  line  into  segments  containing  given  angles. 

43.  To  describe  a  circumference  passing  through  a  given  point,  and 
touching  a  given  line,  or  a  given  circle,  in  a  given  point. 

44.  To  draw  a  tangent  to  a  given  circle,  perpendicular  to  a  given  line. 

45.  Draw  a  line  parallel  to  a  given  line,  so  that  the  segment  intercepted 
between  two  other  given  lines  may  equal  a  given  segment. 

46.  To  draw  a  tangent  to  a  given  circle  which  shall  be  parallel  to 
a  given  straight  line. 

47.  Describe  a  circle  of  given  radius  to  touch  two  given  lines.     Show 
that  the  solution  is,  in  general,  impossible  if  the  lines  are  parallel,  but 
that  otherwise  there  are  four  solutions. 


APPENDIX.  217 


ADDITIONAL   EXERCISES   ON   BOOK   III. 

1.  Any  parallelogram  that  can  be  circumscribed  about  a  circle  must 
be  equilateral. 

2.  Any  parallelogram  that  can  be  inscribed  in  a  circle  will  have  the 
intersection  of  its  diagonals  at  the  centre  of  the  circle. 

3.  The  bisector  of  an  angle  formed  by  a  tangent  and  a  chord  bisects 
the  intercepted  arc. 

4.  Give  the   construction  for  cutting  off   two-sevenths  of   a  given 
straight  line. 

5.  Any  two  altitudes  of  a  triangle  are  inversely  proportional  to  the 
corresponding  bases. 

6.  In  any  isosceles  triangle,  the  square  of  one  of  the  equal  sides  is 
equal  to  the  square  of  any  straight  line  drawn  from  the  vertex  to  the  base 
plus  the  product  of  the  segments  of  the  base. 

7.  The  difference  of  the  squares  of  two  sides  of  any  triangle  is  equal 
to  the  difference  of  the  squares  of  the  projections  of  these  sides  on  the 
third  side. 

8.  If  any  two  chords  cut  within  the  circle,  at  right  angles,  the  sum 
of  the  squares  on  their  segments  equals  the  square  on  the  diameter. 

9.  If  a  straight  line  AB  is  divided  at  C and  D  so  that  AB  x  AD  =  AC2, 
and  if  from  A  any  straight  line  AE  is  drawn  equal  to  AC,  then  EC  bisects 
the  angle  DEB. 

10.  The  intersection  of  the  diagonals  of  an  equiangular  quadrilateral 
is  the  centre  of  the  circumscribed  circle. 

11.  If  two  circles  intersect  in  P,  and  the  tangents  at  P  to  the  two 
circles  meet  the  circles  again  at  Q  and  .R,  prove  that  PQ  :  PR  in  the 
same  ratio  of  the  radii  of  the  circles. 

12.  The  tangents  to  two  intersecting  circles  drawn  from  any  point  in 
their  common  chord  produced,  are  equal. 


218  PLANE  GEOMETEY. 

13.  If  any  two  circles  touch  each  other,  either  internally  or  externally, 
any  two  straight  lines  drawn  from  the  point  of  contact  will  be  cut  pro  • 
portionally  by  the  circumferences. 

14.  If  the  diagonals  of  an  inscribed  quadrilateral  bisect  each  other, 
what  kind  of  a  quadrilateral  is  it  ? 

15.  The  diagonals  of  a  trapezoid  cut  each  other  in  the  same  ratio. 

16.  Describe  a  circumference  passing  through  two  given  points  and 
having  its  centre  in  a  given  straight  line.     When  is  this  impossible  ? 

17.  If  two  circles  touch  each  other,  secants  drawn  through  their  point 
of  contact  and  terminating  in  the  two  circumferences  are  divided  pro- 
portionately at  the  point  of  contact. 

18.  The  intersection  of  the  diagonals  of  an  equilateral  quadrilateral  is 
the  centre  of  the  inscribed  circle. 

19.  If  two  circles  are  tangent  internally,  all  chords  of  the  greater 
circle  drawn  from  the  point  of  contact  are  divided  proportionately  by 
the  circumference  of  the  smaller  circle. 

20.  The  bisectors  of  any  angle  of  an  inscribed  quadrilateral,  and  the 
opposite  exterior  angle,  meet  on  the  circumference. 

21.  ABCD  is  a  quadrilateral  having  two  of  its  sides,  AB,  CD,  par- 
allel.    AF,  CG-  are  drawn  parallel  to  each  other,  meeting  BC,  AD  re- 
spectively in  F,  G.     Prove  that  BG  is  parallel  to  DF. 

22.  If  the  line  of  centres  of  two  circles  meets  the  circumferences  at  the 
points  A,  .B,  (7,  Z>,  and  meets  the  common  exterior  tangent  at  P,  then 
PA  x  PD  =  PB  x  PC. 

23.  Find  a  point  equidistant  from  three  given  points.     When  is  the 
problem  impossible  ? 

24.  A  tree  casts  a  shadow  90  ft.  long,  when  a  vertical  rod  6  ft.  high 
casts  a  shadow  5  ft.  long.     How  high  is  the  tree  ? 

25.  The  sides  of  a  triangle  are  5,  6,  7.     In  a  similar  triangle  the  side 
homologous  to  7  is  equal  to  35.     Find  the  other  two  sides. 

26.  Two  circles  touch  in  (7,  a  point  D  is  taken  outside  them  such  that 
the  radii,  AC,  BC,  subtend  equal  angles  at  Z>,  and  DE,  DF  are  tangents 
to  the  circles.     If  EF  cuts  DG  in  G,  prove  that  DE :  DF  =  EG  :  GF. 


APPENDIX.  219 

27.  The  bisectors  of  the  angles  formed  by  producing  the  opposite  sides 
of  an  inscribed  quadrilateral  to  meet,  are  perpendicular  to  each  other. 

28.  How  long  must  a  ladder  be  to  reach  a  window  24  ft,  high,  if  the 
lower  end  of  the  ladder  is  10  ft.  from  the  side  of  the  house  ? 

29.  If,  in  a  right  triangle,  the  altitude  upon  the  hypotenuse  divides  it 
in  extreme  and  mean  ratio,  the  lesser  arm  is  equal  to  the  farther  segment. 

30.  The  base  of  a  triangle  is  given  and  is  bisected  by  the  centre  of 
a  given  circle.     If  the  vertex  be  at  any  point  of  the  circumference,  show 
that  the  sum  of  the  squares  on  the  two  sides  of  the  triangle  is  constant. 

31.  The  altitudes  of  a  triangle  are  inversely  proportional  to  the  sides 
upon  which  they  are  drawn. 

32.  If  the  diagonals  of  an  inscribed  quadrilateral  are  perpendicular  to 
each  other,  the  line  through  their  intersection  perpendicular  to  any  side 
bisects  the  opposite  side. 

33.  Through  a  given  point  between  two  given  straight  lines,  draw  a 
straight  line  which  shall  be  terminated  by  the  given  lines  and  divided  by 
the  point  in  a  given  ratio. 

34.  The  distance  from  the  centre  of  a  circle  to  a  chord  10  in.  long  is 
12  in.     Find  the  distance  from  the  centre  to  a  chord  24  in.  long. 

35.  The  square  on  the  base  of  an  isosceles  triangle  is  equal  to  twice 
the  product  of  either  side  by  the  part  of  that  side  intercepted  between 
the  perpendicular  let  fall  on  the  side  from  the  opposite  angle  and  the 
end  of  the  base. 

36.  If  two  circles  are  tangent  externally,  a  common  exterior  tangent  is 
a  mean  proportional  between  their  diameters. 

37.  The  radius  of  a  circle  is  6  in.     Through  a  point  10  in.  from  the 
centre  tangents  are  drawn.     Find  the  lengths  of  the  tangents,  and  also  of 
the  chord  joining  the  points  of  contact. 

38.  Upon  a  given  portion  AC  of  the   diameter  AB  of  a  semicircle 
another  semicircle  is  described.      Draw  a  line  through  A  so  that  the 
part  intercepted  between  the  semicircles  may  be  of  given  length. 

39.  If  three  circles  intersect  each  other,  their  three  common  chords 
pass  through  the  same  point. 

40.  Inscribe  a  square  in  a  given  segment  of  a  circle. 


220  PLANE  GEOMETRY. 

41.  From  the  end  of  a  tangent  20  in.  long  a  secant  is  drawn  through 
the  centre  of  the  circle.     If  the  exterior  segment  of  this  secant  is  10  in., 
find  the  radius  of  the  circle. 

42.  From  a  given  point  without  a  circle  draw  a  secant  divided  by  the 
circumference  into  a  given  ratio. 

43.  Divide  any  side  of  a  triangle  into  two  parts  proportional  to  the 
other  sides. 

44.  If  a  perpendicular  is  let  fall  from  any  point  on  the  circumference, 
to  any  diameter,  it  is  the  mean  proportional  between  the  segments  into 
which  it  divides  that  diameter. 

45.  In  a  chord  produced,  find  a  point  such  that  the  tangents  drawn 
from  it  to  the  circle  shall  be  equal  to  a  given  line. 

46.  The  sides  of  a  triangle  are  4,  5,  6.     Is  the  largest  angle  acute,  right, 
or  obtuse  ? 

47.  From  a  given  point  on  the  circumference  of  a  given  circle  draw 
two  chords  so  as  to  be  in  a  given  ratio  and  to  contain  a  given  angle. 


APPENDIX.  221 


ADDITIONAL  EXERCISES   ON   BOOK   IV. 

1.  If  two  triangles  are  on  equal  bases  and  between  the  same  parallels, 
then  any  line  parallel  to  their  bases,  and  cutting  the  triangles,  will  cut 
off  equal  triangles. 

2.  If  the  middle  points  of  two  adjacent  sides  of  a  parallelogram  are 
joined,  a  triangle  is  formed  which  is  equivalent  to  one-eighth  of  the  entire 
parallelogram. 

3.  Two  equilateral  triangles  have  their  areas  in  the  ratio  of  1:2.     Find 
the  ratio  of  their  sides  to  the  nearest  0.01. 

4.  The  sides  of  a  triangle  are  9,  11,  14  inches.     Is  the  triangle  right- 
angled  ?  obtuse-angled  ? 

5.  The  square  of  the  base  of  an  isosceles  triangle  is  equivalent  to  twice 
the  rectangle  contained  by  either  of  the  arms  and  the  projection  of  the 
base  upon  that  side. 

6.  The  perimeter  of  a  rectangle  is  144  ft.,  and  the  length  is  three  times 
the  altitude  ;  find  the  area. 

7.  The  area  of  a  triangle  is  equal  to  the  product  of  its  three  sides 
divided  by  four  times  the  radius  of  the  circumscribed  circle  ;  that  is, 

denoting  this  radius  by  jff, 

c,     abc 


8.  The  area  of  a  trapezoid  is  equal  to  the  product  of  one  of  the  legs 
and  the  distance  from  this  leg  to  the  middle  point  of  the  other  leg. 

9.  Any  quadrilateral  is  divided  by  its  interior  diagonals  into  four 
triangles  which  form  a  proportion. 

10.  The  sides  of  a  triangle  are  4,  11,  13  units  long.     Is  the  angle  oppo- 
site 13  right  ?  obtuse  ?  acute  ? 

11.  Three  times  the  sum  of  the  squares  of  the  sides  of  a  triangle  is 
equal  to  four  times  the  sums  of  the  squares  of  the  medians. 

12.  On  a  given  straight  line  construct  a  triangle  equal  to  a  given  tri- 
angle and  having  its  vertex  in  a  given  straight  line  not  parallel  to  the  base. 

13.  What  part  of  a  parallelogram  is  the  triangle  cut  off  by  a  line 
drawn  from  one  vertex  to  the  middle  point  of  one  of  the  opposite  sides  ? 


222  PLANE  GEOMETRY. 

14.  If  one  angle  of  a  triangle  is  one-third  of  a  straight  angle,  show  that 
the  square  on  the  opposite  side  equals  the  sum  of  the  squares  on  the  other 
two  sides  less  their  rectangle. 

15.  If  any  point  within  a  parallelogram  be  joined  with  the  vertices,  the 
sums  of  the  opposite  pairs  of  triangles  are  equivalent. 

16.  Construct  a  parallelogram  that  shall  be  equal  in  area  and  perimeter 
to  a  given  triangle. 

17.  If  the  side  of  one  equilateral  triangle  is  equal  to  the  altitude  of 
another,  what  is  the  ratio  of  their  areas  ? 

18.  If  two  fixed  parallel  tangents  are  cut  by  a  variable  tangent,  the 
rectangle  of  the  segments  of  the  latter  is  constant. 

19.  Of  the  four  triangles  formed  by  drawing  the  diagonals  of  a  trape- 
zoid,   (1)   those  having  as  bases  the  non-parallel  sides  are  equivalent; 
(2)  those  having  as  bases  the  parallel  sides  are  as  the  squares  of  those 
sides. 

20.  A  triangle  is  divided  by  each  of  its  medians  into  two  parts  of  equal 
area. 

21.  If  two  triangles  have  a  common  angle  and  equal  areas,  the  sides 
containing  the  common  angle  are  inversely  proportional. 

22.  In  AC,  a  diagonal  of  the  parallelogram  ABCD,  any  point  H  is 
taken,  and  HB,  HD  are  drawn  ;  show  that  the  triangle  BAH  is  equal 
in  area  to  the  triangle  DAH. 

23.  The  sum  of  the  squares  of  the  four  segments  of  any  two  chords 
that  intersect  at  right  angles  is  constant. 

24.  If  any  point  in  one  side  of  a  triangle  be  joined  to  the  middle  points 
of  the  other  sides,  the  area  of  the  quadrilateral  thus  formed  is  one-half 
that  of  the  triangle. 

25.  Find  the  ratio  of  a  rectangle  18  yds.  by  14|  yds.  to  a  square  whose 
perimeter  is  100  ft. 

26.  In  a  trapezoid  the  straight  lines,  drawn  from  the  middle  points  of 
one  of  the  non-parallel  sides  to  the  ends  of  the  opposite  sides,  form  with 
that  side  a  triangle  equal  to  half  the  trapezoid. 

27.  Show  that  the  line  joining  the  middle  points  of  the  two  parallel 
sides  of  a  trapezoid  divides  the  area  into  two  equal  parts. 

28.  To  transform  a  parallelogram  into  a  parallelogram  having  one  side 
equal  to  a  given  length. 

29.  Construct  an  isosceles  triangle  on  the  same  base  as  a  given  triangle, 
and  equivalent  to  it. 


APPENDIX.  223 

30.  Construct  a  parallelogram  having  a  given  angle  upon  the  same 
base  as  a  given  square,  and  equivalent  to  it. 

31.  To  construct  a  triangle  equal  to  a  given  parallelogram,  and  having 
one  of  its  angles  equal  to  a  given  angle. 

32.  Construct  an  isosceles  triangle  equal  in  area  to  a  given  triangle 
and  having  a  given  vertical  angle. 

33.  To  find  a  point  within  a  triangle,  such  that  the  lines  joining  this 
point  to  the  vertices  shall  divide  the  triangle  into  three  equivalent  parts. 

34.  Divide  a  given  line  into  two  segments,  such  that  their  squares  shall 
be  as  8  :  5. 

35.  On  a  given  line  to  construct  a  rectangle  equal  to  a  given  rectangle. 

36.  With  a  given  altitude  to  construct  an  isosceles  triangle  equal  to  a 
given  triangle. 

37.  Divide  a  straight  line  into  two  parts,  such  that  the  sum  of  the 
squares  on  the  parts  may  be  equal  to  a  given  square. 

38.  To  divide  a  given  triangle  into  two  equivalent  parts  by  drawing  a 
line  perpendicular  to  one  of  the  sides. 

39.  To  construct  a  triangle,  given  its  angles  and  its  area. 

40.  Bisect  a  triangle  by  a  line  parallel  to  the  base. 

41.  On  the  base  of  a  given  triangle  to  construct  a  rectangle  equal  to 
the  given  triangle. 

42.  Construct  a  square  that  shall  be  one  third  of  a  given  square. 

43.  Given  any  triangle,  to  construct  an  isosceles  triangle  of  the  same 
area  whose  vertical  angle  is  an  angle  of  the  given  triangle. 

44.  To  construct  a  square  equal  to  half  the  sum  of  two  given  squares. 

45.  Construct  a  parallelogram  equal  to  a  given  triangle  and  having  one 
of  its  angles  equal  to  a  given  angle. 

46.  Construct  a  triangle  equivalent  to  a  given  triangle,  and  having  one 
side  equal  to  a  given  line. 

47.  Construct  a  triangle  similar  to  a  given  triangle  ABC  which  shall  be 
to  ABC  in  the  ratio  of  AB  to  BC. 

48.  Construct  a  triangle  equal  to  a  given  triangle  and  having  one  of  its 
angles  equal  to  an  angle  of  the  triangle,  and  the  sides  containing  this 
angle  in  a  given  ratio. 

49.  Construct  a  square  that  shall  be  to  a  given  triangle  as  7  is  to  6. 

50.  Bisect  a  triangle  by  a  straight  line  drawn  through  a  given  point 
in  one  of  its  sides. 


224  PLANE  GEOMETRY. 


ADDITIONAL   EXERCISES   ON   BOOK  V. 

1.  What  is  the  radius  of  the  circle  circumscribing  the  triangle  whose 
sides  are  3,  4,  5  ? 

2.  The  area  of  the  regular  inscribed  hexagon  is  half  the  area  of  the 
circumscribed  equilateral  triangle. 

3.  The  apothem  of  a  regular  pentagon  is  6  and  a  side  is  4  ;  find  the 
perimeter  and  area  of  a  regular  pentagon  whose  apothem  is  8. 

4.  The  area  of  an  inscribed  regular  hexagon  is  a  mean  proportional 
between  the  areas  of  the  inscribed   and  the   circumscribed  equilateral 
triangles. 

5.  The  area  of  an  inscribed  regular  octagon  is  equal  to  that  of  a 
rectangle  whose  sides  are  equal  to  the  sides  of  the  inscribed   and  the 
circumscribed  squares. 

6.  If  the  diagonals  of  an  inscribed  quadrilateral  are  perpendicular  to 
each  other,  then  the  sum  of  the  products  of  the  two  opposite  sides  equals 
twice  the  area  of  the  quadrilateral. 

7.  The  apothem  of  an  inscribed  equilateral  triangle  is  equal  to  half 
the  radius  of  the  circle. 

8.  The  radius  of  a  circle  is  8  ;  find  the  apothem,  perimeter,  and  area 
of  the  inscribed  equilateral  triangle. 

9.  If  a  =  the  side  of  a  regular  pentagon  inscribed  in  a  circle  whose 
radius  is  It,  then, 


10.  If  a  =  the  side  of  a  regular  octagon  inscribed  in  a  circle  whose 
radius  is  R,  then, 

a  =  It  V2  -  V2. 

11.  Upon  the  six  sides  of  a  regular  hexagon  squares  are  constructed 
outwardly.     Prove  that  the  exterior  vertices  of  these  squares  are  the 
vertices  of  a  regular  dodecagon. 

12.  The  area  of  an  inscribed   equilateral  triangle  is  half  that  of   a 
regular  hexagon  inscribed  in  the  same  circle. 


APPENDIX.  225 

13.  If  a  =  the  side  of  a  regular  dodecagon  inscribed  in  a  circle  whose 
radius  is  7?,  then,  _ 

a  =  It  V2  -  V3. 

14.  The  radius  of  a  circle  is  ten  :  find  the  perimeter  and  area  of  the 
regular  inscribed  octagon. 

The  radius  of  a  circle  is  4  ;  find  the  area  of  the  inscribed  square. 

15.  The  radius  of  an  inscribed  regular  polygon  is  the  mean  propor- 
tional between  its  apothem  and  the  radius  of  the  similar  circumscribed 
polygon. 

16.  What  is  the  radius  of  that  circle  of  which  the  number  of  square 
units  of  area  equals  the  number  of  linear  units  of  circumference  ? 

17.  The  altitude  of  an  equilateral  triangle  is  to  the  radius  of  the  cir- 
cumscribing circle  as  3  is  to  2. 

18.  If   a  =  the  side  of   a  regular  pentedecagon  inscribed  in  a  circle 
whose  radius  is  7?,  then, 


a  =  J?  (  VlO  +  2  V5  +  V3  -  Vl5). 
4 

19.  The  chord  of  an  arc  is  24  in.,  and  the  height  of  the  arc  is  9  in. 
Find  the  diameter  of  the  circle. 

20.  What  is  the  radius  of  that  circle  of  which  the  number  of  square 
units  of  area  equals  the  number  of  linear  units  of  radius  ? 

21.  The  square  inscribed  in  a  semicircle  is  to   that  inscribed  in  a 
circle  as  2  is  to  5. 

22.  The  area  of  the  regular  inscribed  hexagon  is  equal  to  twice  the 
area  of  the  regular  inscribed  triangle. 

23.  The  diagonals  drawn  from  the  vertex  of   a  regular  pentagon  to 
the  opposite  vertices  trisect  that  angle. 

24.  If  a  =  the  side  of  a  regular  polygon  in  a  circle  whose  radius  is  72, 
and  A  =  the  side  of  the  similar  circumscribed  polygon,  then, 


25.  Find  the  area  of  a  sector,  if  the  radius  of  the  circle  is  28  ft.,  and 
the  angle  at  the  centre  45°. 

26.  Find  the  areas  of  circles  with  radii  5,  8,  21,  33,  47,  52.     (In  these 
computations,  let  IT  =  3.1416.) 

27.  The  intersecting  diagonals  of  a  regular  pentagon  divide  each  other 
in  extreme  and  mean  ratio. 


226  PLANE   GEOMETRY. 

28.    If   d  -  the   diagonal  of   a  regular  pentagon  inscribed  in  a  circle 
whose  radius  is  _K,  then, 


29.  The  square  of  the  side  of  the  inscribed  equilateral  triangle  is  three 
times  the  square  of  a  side  of  the  regular  inscribed  hexagon. 

30.  The  perpendiculars  from  two  vertices  of  a  triangle  upon  the  oppo- 
site sides  divide  each  other  into  segments  reciprocally  proportional. 

31.  Find  the  areas  of  circles  with  diameters  2,  8,  11,  31,  42,  97. 

32.  Inscribe  an  equilateral  triangle  in  a  given  square,  so  as  to  have 
a  vertex  of  the  triangle  at  a  vertex  of  the  square. 

33.  The  area  of  a  triangle  is  equal  to  half  the  product  of  its  perimeter 
by  the  radius  of  the  inscribed  circle. 

34.  The  Egyptians  said:    "Construct  a  square  the  side  of   which  is 
|  of  the  diameter  of  a  circle,  and  its  area  will  equal  that  of  the  circle." 
From  this  compute  their  value  of  TT. 

35.  Construct  a  square  that  shall  be  f  of  a  given  square. 

36.  The  square  inscribed   in   a  semicircle  is  equal  to  -f  the  square 
inscribed  in  the  whole  circle. 

37.  Lines  drawn  from  one  vertex  of   a  parallelogram  to  the  middle 
points  of  the  opposite  sides  trisect  one  of  the  diagonals. 

38.  Of  all  triangles  in  a  given   circle,  that  which   has  the  greatest 
perimeter  is  equilateral. 

39.  Construct  a  regular   hexagon  that  shall  be  f  of  a  given  regular 
hexagon. 

40.  The  area  of  a  circle  is  40  ft.  ;  find  the  side  of  the  inscribed  square. 

41.  Construct  a  square  equivalent  to  the  sum  of  a  given  triangle  and 
a  given  parallelogram. 

42.  Find  a  point  in  a  given  straight  line  such  that  the  tangents  drawn 
from  it  to  a  given  circle  contain  the  maximum  angles. 

43.  Find  the  angle  subtended  at  the  center  of  a  circle  by  an  arc  6  ft. 
long,  if  the  radius  is  8  ft.  long. 

44.  To  construct  a  triangle,  given  its  angles  and  its  area. 

45.  Through  a  point  of  intersection  of  two  circumferences,  draw  the 
maximum  line  terminated  by  the  two  circumferences. 

46.  Find  the  length  of  the  arc  subtended  by  one  side  of  a  regular 
octagon  inscribed  in  a  circle  whose  radius  is  10  ft. 


APPENDIX.  227 

47.  To  construct  an  equilateral  triangle  having  a  given  area. 

48.  Of  all  triangles  of  a  given  base  and  area  the  isosceles  has  the 
greatest  vertical  angle. 

49.  Find  the  area  of  a  circular  sector,  the  chord  of  half  the  arc  being 
10  in.  and  the  radius  25  in. 

50.  What  is  the  area  of  the  largest  triangle  that  can  be  inscribed  in 
a  circle  of  radius  10  ? 

51.  To  construct  a  triangle,  given  its  base,  the  ratio  of  the  other  sides, 
and  the  angle  included  by  them. 

52.  The  radius  of  a  circle  is  5  ft.    Find  the  radius  of  a  circle  16  times 
as  large. 

53.  Every  equilateral  polygon  circumscribed  about  a  circle  is  equi- 
angular, if  the  number  of  sides  be  odd. 

54.  Given  a  square  of  area  1.      Find   the   area  of  an  isoperhnetric 
(1)  equilateral  triangle,  (2)  regular  hexagon,  (3)  circle. 

55.  Find  the  height  of  an  arc,  the  chord  of  half  the  arc  being  10  ft., 
and  the  radius  24  ft. 

56.  What  is  the  only  rectilinear  polygon  that  is  necessarily  plane  ? 
Why  ? 

57.  Find  the  length  of  the  arc  subtended  by  one  side  of  a  regular 
dodecahedron  in  a  circle  whose  radius  is  12.5  ft. 

58.  Find  the  area  of  a  segment  whose  height  is  16  in.,  the  radius  of 
the  circle  being  20  in. 

59.  Find  the  area  of  a  segment  whose  arc  is  100,  the  radius  being  24  ft. 

60.  If  AB  be  a  side  of  an  equilateral  triangle  inscribed  in  a  circle,  and 
AD  a  side  of  the  inscribed  square,  prove  that  three  times  the  square  on 
AD  is  equal  to  twice  the  square  on  AB. 


228  PLANE  GEOMETRY. 


MISCELLANEOUS   QUESTIONS.     BOOKS   I.-V. 

(PLANE  GEOMETRY.) 

1.  The  sum  of  the  distances  of  any  point  in  the  base  of  an  isosceles 
triangle  from  the  equal  sides  is  equal  to  the  distance  of  either  extremity 
of  the  base  from  the  opposite  side. 

2.  Prove  that  the  square  constructed  on  the  difference  of  two  straight 
lines  is  equal  to  the  sum  of  the  squares  constructed  on  the  lines,  dimin- 
ished by  twice  the  rectangle  of  the  lines. 

3.  Any  chord  of  a  circle  is  a  mean  proportional  between  its  projection 
on  the  diameter  from  any  one  of  the  extremities,  and  the  diameter  itself. 

4.  To  divide  a  circle  into  two  segments  so  that  the  angle  contained  in 
one  shall  be  double  that  contained  in  the  other. 

5.  The  bisector  of  an  exterior  angle  at  the  vertex  of  an  isosceles  tri- 
angle is  parallel  to  the  base. 

6.  The  bisectors  of  the  external  angles  of  a  quadrilateral  form  a  cir- 
cumscribed quadrilateral,  the  sum  of  whose  opposite  angles  is  equal  to 
two  right  angles. 

7.  The  angles  made  with  the  base  of  an  isosceles  triangle  by  perpen- 
diculars from  its  extremities  on  the  equal  sides  are  each  equal  to  half  the 
vertical  angle. 

8.  Construct  a  triangle,  having  given  the  base,  the  vertical  angle,  and 
(1)  the  sum,  or  (2)  the  difference  of  the  sides. 

9.  Given  the  base,  one  of  the  angles  at  the  base,  and  the  difference 
of  the  sides  of  a  triangle,  to  construct  the  triangle. 

10.  Given  two  sides  of  a  triangle  and  the  straight  line  drawn  from  the 
extremity  of  one  of  them  to  the  middle  point  of  the  other,  to  construct 
the  triangle. 

11.  In  the  triangle  ABC,  the  angle  A  =  50°,  the  angle  B  =  70°.     What 
angle  will  the  bisectors  of  these  two  angles  make  with  each  other  ? 

12.  How  many  sides  has  a  polygon,  the  sum  of  whose  interior  angles 
is  four  times  that  of  its  exterior  angles  ? 


APPENDIX.  229 

13.  In  a  given  circle  to  draw  a  chord  equal  and  parallel  to  a  given  line. 

14.  From  a  given  isosceles  triangle  cut  off  a  trapezoid  having  for  base 
that  of  the  triangle,  and  having  its  other  three  sides  equal. 

15.  Find  the  number  of  degrees  in  the  arc  whose  length  is  equal  to  the 
radius  of  the  circle. 

16.  A  straight  line  touches  a  circle  at  A,  and  from  any  point  P,  in  the 
tangent,  PB  is  drawn  meeting  the  circle  at  B  so  that  PB  =  PA.     Prove 
that  PB  touches  the  circle. 

17.  If  one  of  the  parallel  sides  of  a  trapezoid  is  double  the  other,  the 
diagonals  intersect  one  another  in  a  point  of  trisection. 

* 

18.  Find  the  side  of  a  square  equivalent  to  a  circle  whose  radius  is  40  ft. 

19.  Find  the  radius  of  the  circle  whose  sector  of  45°  is  .125  sq.  in. 

20.  A  circle  is  described  passing  through   the  ends  of  the  base  of  a 
given  triangle  ;  prove  that  the  straight  line  joining  the  points,  in  which 
it  meets  the  sides  or  the  sides  produced,  is  parallel  to  a  fixed  straight  line. 

21.  Two  circles  touch   externally  at  A ;   the  tangent  at  B  to  one  of 
them  cuts  the  other  in  C,  D;  prove  that  BG  and  BD  subtend  supple- 
mentary angles  at  A. 

22.  C  is  the  centre  of  a  given  circle,   CA  a  radius,  B  a  point  on  a 
radius  at  right  angles  to  CA  ;  join  AB  and  produce  it  to  meet  the  circle 
again  at  Z>,  and  let  the  tangent  at  D  meet  CB  produced  at  E :  show  that 
BDE  is  an  isosceles  triangle. 

23.  What  is  the  width  of  a  ring  between  two  concentric  circumferences 
whose  lengths  are  160  ft.  and  80  ft.  ? 

24.  The  circumference  of  a  circle  is  78.54  in.  ;  find  (1)  its  diameter, 
and  (2)  its  area. 

25.  Find  the  point  inside  a  given  triangle  at  which  the  sides  subtend 
equal  angles. 

26.  The  figure  formed  by  the  five  diagonals  of  a  regular  pentagon  is 
a  regular  pentagon. 

27.  With  a  given  radius,  describe  a  circle  touching  two  given  circles. 

28.  Describe  a  circumference  passing  through  a  given  point  and  touch- 
ing a  given  line  in  a  given  point. 

29.  Through   one1  of   the   points  of   intersection  of  two  given   circles 
draw  a  secant  forming  chords  that  are  in  a  given  ratio. 

30.  Describe  a  circumference  touching  two  parallel  lines  and  passing 
through  a  given  point. 


230  PLANE   GEOMETRY. 

31.  If  three  circles  touch  one  another  externally  in  P,  Q,  .#,  and  the 
chords  PQ,  PR  of  two  of  the  circles  be  produced  to  meet  the  third  circle 
again  in  ST,  prove  that  ST  is  a,  diameter. 

32.  Two  tangents  are   drawn   to  a  circle  at  the  opposite  extremities 
of  a  diameter,  and  intercept  from  a  third  tangent  a  portion  AB ;  if  C 
be  the  centre  of  the  circle,  show  that  ACE  is  a  right  angle. 

33.  Through  the  vertices  of  a  quadrilateral  straight  lines  are  drawn 
parallel  to  the  diagonals ;  prove  that  the  figure  thus  formed  is  a  paral- 
lelogram which  is  double  the  quadrilateral. 

34.  Describe  a  circle  which  shall  pass  through  two  given  points  and 
touch  a  given  straight  line.     Two  solutions. 

35.  To  divide  one  side  of  a  given  triangle  into  segments  proportional 
to  the  adjacent  sides. 

36.  To   describe  a  circle  which  shall  pass  through   two  given  points 
and  touch  a  given  circle. 

37.  Show  that  the  sum  of  the  perpendiculars  from  any  point  inside 
a  regular  hexagon  to  the  six  sides  is  equal  to  three  times  the  diameter 
of  the  inscribed  circle. 

38.  The  three  sides  of  a  triangle  are  9,  10,  17  in.,  respectively  ;  find 
(1)  its  area  and  (2)  the  area  of  the  inscribed  circle. 

39.  In  a  given  circle  inscribe  a  triangle  similar  to  a  given  triangle. 

40.  Through  one  of  the  points  of  intersection  of  two  circumferences, 
draw  a  straight  line,  terminated  by  the  circumferences,  which  shall  have 
a  given  length. 

41.  Construct  a  parallelogram,  having  given   (1)    two   diagonals  and 
the  angle  between  them,  (2)  one  side,  one  diagonal,  and  the  angle  be- 
tween the  diagonals. 

42.  Describe  a  circle  with  given  radius  to  touch  a  given  line  in  a  given 
point.     How  many  such  circles  can  be  described  ? 

43.  Construct  a  triangle,  having  given  a  median  and  the  two  angles 
into  which  the  angle  is  divided  by  that  median. 

44.  Every  equiangular  polygon  inscribed  in  a  circle  is  equilateral   if 
the  number  of  sides  be  odd. 

45.  Prove  that  the  rectangle  of  the  sum  and  difference  of  two  straight 
lines  is  equal  to  the  difference  of  the  squares  constructed  on  the  lines. 

46.  If  the  straight  line  joining  the  middle  points  of  two  opposite  sides 
of  any  quadrilateral  divide  the  area  into  two  equal  parts,  show  that  the 
two  bisected  sides  are  parallel. 


APPENDIX.  231 

47.  Describe  a  circle  which  shall  touch  a  given  straight  line  at  a  given 
point  and  pass  through  another  given  point  not  in  the  line. 

48.  The  apothem  of   a  regular   hexagon  is  12  ;   find   the  area  of  the 
circumscribing  circle. 

49.  The  angle  included  between  the  internal  bisector  of  one  base  angle 
of  a  triangle  and  the  external  bisector  of  the  other  base  angle  is  equal 
to  half  the  vertical  angle. 

50.  If  the  exterior  angles  of  a  triangle  are  bisected,  the  three  exterior 
triangles  formed  on  the  sides  of  the  original  triangle  are  equiangular, 

51.  The  angle  formed  by  the  bisectors  of  any  two  consecutive  angles 
of  a  quadrilateral  is  equal  to  the  sum  of  the  other  two  angles. 

52.  AB  is  the  diameter  and  C  the  centre  of  a  semicircle  ;  show  that 
0,  the  centre  of   any  circle  inscribed   in   the   semicircle,  is   equidistant 
from  C  and  from  the  tangent  to  the  semicircle  parallel  to  AB. 

53.  To  find  in  one  side  of  a  given  triangle  a  point  whose  distances 
from  the  other  sides  shall  be  to  each  other  in  a  given  ratio. 

54.  Through  a  point  in  a  circle  draw  a  chord  that  is  bisected  in  that 
point,  and  show  that  it  is  the  least  chord  through  that  point. 

55.  To  draw  through  a  point  P,  exterior  to  a  given  circle,  a  secant 
PAB  so  that  AP  :  BP  =2:3. 

56.  Prove  that  the  square  constructed  on  the  sum  of  two  straight  lines 
is  equal  to  the  sum  of  the  squares  upon  each  of  the  two  straight  lines 
plus  twice  the  rectangle  of  the  lines. 

57.  Having  given  the  greater  segment  of   a  line   divided  in  extreme 
and  mean  ratio,  to  construct  the  line. 

58.  Construct  a  right  triangle,  having  given  the  hypotenuse  and  the 
perpendicular  from  the  right  angle  on  it. 

59.  The  position  and  magnitude  of  two  chords  of  a  circle  being  given, 
describe  the  circle. 

60.  Construct  a  right  triangle,  having  given  the  hypotenuse  and  the 
difference  of  the  other  sides. 

61.  If  one  angle  of  a  triangle  is  equal  to  the  sum  of  the  other  two,  the 
triangle  can  be  divided  into  two  isosceles  triangles. 

62.  BAG  is  a  triangle  having  the  angle  B  double  the  angle  A.     If  BD 
bisects  the  angle  B  and  meets  AC  at  Z>,  show  that  BD  is  equal  to  AD. 


232  PLANE  GEOMETRY. 

63.  ABC,  DEF  are  triangles  having  the  angles  A  and  D  equal,  and 
AB  equal  to  DE.     Prove  that  the  triangles  are  to  each  other  as  AC 
is  to  DF. 

64.  Construct  a  parallelogram,  having  given  : 

Two  adjacent  sides  and  a  diagonal. 
A  side  and  both  diagonals. 

65.  With  a  given  point  as  centre  describe  a  circle  which  shall  intersect 
a  given  circle  at  the  ends  of  a  diameter. 

66.  Through  a  given  point  within  a  given  circle  draw  two  equal  chords 
which  shall  contain  a  given  angle. 

67.  Through  a  given  point  inside  the  circle  which  is  not  the  centre, 
draw  a  chord  bisected  at  that  point. 


APPENDIX.  233 


ADDITIONAL   EXERCISES   ON   BOOK  VI. 

1.  What  is  the  reason  that  a  three-legged  chair  is  always  stable  on 
the  floor  while  a  four-legged  one  may  not  be  ? 

2.  Prove  that  parallel  lines  have  their  projections  on  the  same  plane 
in  lines  that  are  coincident  or  parallel. 

3.  Show  that  all  the  propositions  in  Plane  Geometry  which  relate  to 
triangles  are  true  of  triangles  in  space,  however  situated. 

4.  Show  that  those  propositions  are  not  true  of  polygons  of  more  than 
three  sides  situated  in  any  way  in  space. 

5.  To   construct  a  plane   containing  a  given   line,    and   parallel  to 
another  given  line. 

6.  If  the  projections  of  a  number  of  points  on  a  plane  are  in  a  straight 
line,  these  points  are  in  one  plane. 

7.  A  plane  can  be  passed  perpendicular  to  only  one  edge  or  to  two 
faces  of  a  polyedral  angle. 

8.  If  each  of  the  projections  of  the  line  AB  upon  two  intersecting 
planes  is  a  straight  line,  the  line  AB  is  a  straight  line. 

9.  The  edge  of  a  diedral  angle  is  perpendicular  to  the  plane  of  the 
measuring  angle. 

10.  If  a  line  makes  equal  angles  with  three  lines  in  the  same  plane,  it 
is  perpendicular  to  that  plane. 

11.  If  a  plane  bisects  a  line  perpendicularly,  every  point  of  the  plane 
is  equally  distant  from  the  extremities  of  the  line. 

12.  Through  a  given  point,  to  pass  a  plane  perpendicular  to  a  given 
straight  line. 

13.  Through  a  given  straight  line,  to  pass  a  plane  perpendicular  to 
a  given  plane. 

14.  The  faces  of  a  diedral  angle  are  perpendicular  to  the  plane  of  the 
measuring  angle. 

15.  If  a  plane  be  passed  through  one  diagonal  of  a  parallelogram,  the 
perpendiculars  to  that  plane  from  the  extremities  of  the  other  diagonal 
are  equal. 


234  SOLID    GEOMETRY. 

16.  If  four  lines  in  space  are  parallel,  in  how  many  planes  may  they 
lie  when  taken  two  at  a  time  ? 

17.  To  bisect  a  diedral  angle. 

18.  Through  a  given  line  in  a  plane  pass  a  plane  making  a  given  angle 
with  that  plane. 

19.  If  two  lines  not  in  the  same  plane  are  intersected  by  the  same  line, 
how  many  planes  may  be  determined  by  the  three  lines  taken  two  and  two  ? 

20.  Through  a  given  point,  to  pass  a  plane  parallel  to  a  given  plane. 

21.  If  two  parallel  planes  intersect  two  other  parallel  planes,  the  four 
lines  of  intersection  are  parallel. 

22.  Through  the  edge  of  a  given  diedral  angle  pass  a  plane  bisecting 
that  angle. 

23.  To  draw  a  straight  line  perpendicular  to  a  given  plane  from  a  given 
point  outside  of  it. 

24.  To  draw  a  straight  line  perpendicular  to  a  given  plane  from  a  given 
point  in  the  plane. 

25.  To  determine  that  point  in  a  given  straight  line  which  is  equi- 
distant from  two  given  points  not  in  the  same  plane  with  the  given  line. 

26.  Two  parallel  planes  intersecting  two  parallel  lines  cut  off  equal 
segments. 

27.  In  a  given  plane  find  a  point  equidistant  from  three  given  points 
without  the  plane. 

28.  Parallel  lines  make  equal  angles  with  parallel  planes. 

29.  A  straight  line  makes  equal  angles  with  parallel  planes. 

30.  If  a  line  is  parallel  to  each  of  two  intersecting  planes,  it  is  parallel 
to  their  intersection. 

31.  If  a  line  is  parallel  to  each  of  two  planes,  the  intersections  which 
any  plane  passing  through  it  makes  with  the  planes  are  parallel. 

32.  To  determine  the  point  whose  distances  from  the  three  faces  of 
a  given  triedral  angle  are  given.     Is  it  unique  ? 

33.  If  the  projections  of  any  line  upon  two  intersecting  planes  are  each 
of  them  straight  lines,  prove  that  the  line  itself  is  a  straight  line. 

34.  Two  planes  which  are  not  parallel  are  cut  by  two  parallel  planes. 
Prove  that  the  intersections  of  the  first  two  with  the  last  two  contain 
equal  angles. 

35.  Pass  a  plane  through  a  given  point  parallel  to  a  given  plane. 


APPENDIX.  235 


ADDITIONAL   EXERCISES   ON   BOOK   VII. 

1.  The  lateral  surface  of  a  pyramid  is  greater  than  the  base. 

2.  The  lateral  area  of  a  cylinder  of  revolution  is  equal  to  the  area 
of  a  circle  whose  radius  is  a  mean  proportional   between  the  altitude 
and  diameter  of  the  cylinder. 

3.  An  open  cistern  6  ft.  long  and  4|  ft.  wide  holds  108  cubic  ft.  of 
water.     How  many  cubic  feet  of  lead  will  it  take  to  line  the  sides  and 
bottom,  if  the  lead  is  \  in.  thick  ? 

4.  Find  the  surface  and  volume  of  a  rectangular  parallelepiped  whose 
edges  are  4,  5,  and  6  ft. 

5.  What  is  the  volume  of  a  right  prism  whose  altitude  is  45  in.  and 
whose  base  contains  3  sq.  ft.  ? 

6.  Find  the  volume  of  a  rectangular  parallelepiped  whose  surface  is 
104  and  whose  base  is  2  by  6. 

7.  How  many  square  feet  of  lead  will  be  required  to  line  a  cistern 
open  at  the  top,  which  is  4  ft.  6  in.  long,  2  ft.  8  in.  wide,  and  contains 
42  cu.  ft.  ? 

8.  A  brick  has  the  dimensions,  25  cm.,  12  cm.,  6  cm.,  but  on  account 
of  shrinkage  in  baking,  the  mould  is  27.5  cm.  long  and  proportionally 
wide  and  deep.     What  per  cent  does  the  volume  of  the  brick  decrease 
in  baking? 

9.  The  altitude  of  a  pyramid  is  divided  into  five  equal  parts  by  planes 
parallel  to  the  base.     Find   the  ratios  of   the  various  frustums  to  one 
another  and  to  the  whole  pyramid. 

10.  To  find  two  straight  lines  in  the  ratio  of  the  volumes  of  two  given 
cubes. 

11.  To  cut  a  cube  by  a  plane  so  that  the  section  shall  be  a  regular 
hexagon. 

12.  The  lateral  areas  of   the  two  cylinders  generated  by  revolving  a 
rectangle  successively  about  each  of  its  containing  sides  are  equal. 

13.  Find  the  volume  of  a  cube  the  diagonal  of  whose  face  is  a  \/2. 


236  SOLID    GEOMETRY. 

14.  The  dimensions  of  one  rectangular  parallelepiped  are  2  ft.,  15  ft., 
and   14  ft.,  respectively  ;   those  of  another  are  4  ft.,  5  ft.,  and  13  ft., 
respectively.     What  is  the  ratio  of  the  first  solid  to  the  second  ? 

15.  Find  the  length  of   the  diagonal  of  a  rectangular  parallelepiped 
whose  edges  are  4,  5,  and  6. 

16.  The  base  of  a  pyramid  contains  169  sq.  ft.     A  plane  parallel  to  the 
base  and  4  ft.  from  the  vertex  cuts  a  section  containing  64  sq.  ft.  ;  find 
the  height  of  the  pyramid. 

17.  If  a  slant  height  of  a  cone  of  revolution  is  equal  to  the  diameter  of 
its  base,  its  total  area  is  to  that  of  the  inscribed  sphere  as  9  is  to  4. 

18.  What  should  be  the  edge  of  a  cubical  box  that  shall  contain  8 
gallons  dry  measure  ? 

19.  Find  the  volume  of  a  rectangular  parallelepiped  whose  surface  is 
832  and  whose  base  is  8  by  6. 

20.  The  dimensions  of  a  trunk  are  5  ft.,  4  ft.,  3  ft.     What  are  the 
dimensions  of  a  similar  trunk  holding  four  times  as  much  ? 

21.  Find  the  difference  between  the  volume  of  the  frustum  of  a  pyramid 
and  the  volume  of  a  prism  of  the  same  altitude,  whose  base  is  a  section  of 
the  frustum  parallel  to  its  bases  and  equidistant  from  them. 

The  difference  may  be  expressed  in  the  form 


12 

if  B  and  b  are  the  areas  of  the  bases,  and  h  the  altitude  of  the  frustum. 

22.  A  pyramid  24  ft.  high  has  a  square  base  measuring  16  ft.  on  a  side. 
What  will  be  the  area  of  a  section  made  by  a  plane  parallel  to  the  base 
and  4  ft.  from  the  vertex  ? 

23.  An  equilateral  triangle  revolves  about  one  of  its  altitudes.     What  is 
the  ratio  of  the  lateral  surface  of  the  generated  cone  to  that  of  the  sphere 
generated  by  the  circle  inscribed  in  the  triangle  ? 

24.  What  should  be  the  edge  of  a  cube  so  that  its  entire  surface  shall 
be  2  sq.  ft.  ? 

25.  The  height  of  a  regular  hexagonal  pyramid  is  36  ft.,  and  one  side 
of  the  base  is  6  ft.     What  are  the  dimensions  of  a  similar  pyramid  whose 
volume  is  one-twentieth  that  of  the  first  ? 

26.  A  man  wishes  to  make  a  cubical  cistern  whose  contents  are  186,624 
cu.  in.  ;  how  many  feet  of  inch  boards  will  line  it  ? 

27.  Find  the  height  in  feet  of  a  pyramid  when  the  volume  is  26  cu.  ft. 
936  cu.  in.  ,  and  each  side  of  its  square  base  is  3  ft.  6  in. 


APPENDIX.  237 

28.  The  base  of  a  pyramid  is  18  sq.  ft.  and  its  altitude  is  9  ft.     What  is 
the  area  of  a  section  parallel  to  the  base  and  3  ft.  from  it  ? 

29.  A  conical  tent  of  slant  height  10  ft.  covers  a  circular  area  10  ft.  in 
diameter.     Find  the  volume,  and  the  area  of  canvas. 

30.  The  lateral  edge  of  a  right  prism  is  equal  to  the  altitude. 

31.  The  base  edge  of  a  regular  pyramid  with  a  square  base  is  40  ft., 
the  lateral  edge  101  ft.  ;  find  its  volume  in  cubic  feet. 

32.  The  total  area  of  the  equilateral  cylinder  inscribed  in  a  sphere  is  a 
mean  proportional  between  the  area  of  the  sphere  and  the  total  area  of 
the  inscribed  equilateral  cone.     The  same  is  true  of  the  volumes  of  these 
bodies. 

33.  The  volumes  of  two  similar  cones  are  54  cu.  ft.  and  432  cu.  ft. 
The  height  of  the  first  is  6  ft. ,  what  is  the  height  of  the  other  ? 

34.  The  base  of  a  cone  is  equal  to  a  great  circle  of  a  sphere,  and  the 
altitude  is  equal  to  a  diameter  of  the  sphere.     What  is  the  ratio  of  their 
volumes  ? 

35.  The  perimeter  of  the  base  of  a  pyramid  is  20  in.  ;  its  slant  height 
is  9  in.     What  is  the  lateral  surface  ? 

36.  Find  the  dimensions  of  a  right  circular  cylinder  fifteen-sixteenths 
as  large  as  a  similar  cylinder  whose  height  is  40  ft.,  and  diameter  20  ft. 

37.  The  lateral  areas  of  right  prisms  of  equal  altitudes  are  as  the 
perimeters  of  their  bases. 

38.  The  volume  of  a  sphere  is  to  that  of  the  inscribed  cube  as  TT  is 
to  2  -  V3- 

39.  The  height  of  a  frustum  of  a  right  cone  is  two-fifths  the  height  of 
the  entire  cone.     Compare  the  volumes  of  the  frustum  and  the  entire 
cone. 

40.  The  bases  of  two  pyramids  are  8.1  sq.  ft.  and  10  sq.  ft.,  respec- 
tively ;  their  altitudes  are  10  ft.  and  9  ft.  respectively.     What  is  their 
ratio  ? 

41.  Having  the  base  edge  a,  and  the  total  surface    T7,  of  a  regular 
pyramid  with  a  square  base,  find  the  volume  V. 

42.  If  the  lateral  surface  of  a  right  circular  cylinder  is  a,  and  the  vol- 
ume is  &,  find  the  radius  of  the  base  and  the  height. 

43.  If  the  four  diagonals  of  a  quadrangular  prism  pass  through  a  com- 
mon point,  the  prism  is  a  parallelepiped. 

44.  A  sphere  is  to  the  circumscribed  cube  as  TT  is  to  6. 


238  SOLID    GEOMETRY. 

45.  The  bases  of  a  frustum  of  a  pyramid  are  9  sq.  ft.  and  5J  sq.  ft. 
respectively,  and  its  altitude  is  6  ft.     What  is  its  volume  ? 

46.  The  height  of  a  right  circular  cone  is  equal  to  the  diameter  of  its 
base  ;  find  the  ratio  of  the  area  of  the  base  to  the  lateral  surface. 

47.  Any  straight  line  drawn  through  the  centre  of  a  parallelepiped, 
terminating  in  a  pair  of  faces,  is  bisected  at  the  centre. 

48.  The  bases  of  a  frustum  of  a  pyramid  are  24  sq.  in.  and  8.3  sq.  in. 
respectively.     Its  volume  is  500  cu.  in.     What  is  its  altitude  ? 

49.  Find  the  volume  of  a  prism  the  area  of  whose  base  is  24  sq.  in.  and 
altitude  7  ft. 

50.  Every  section  of  a  prism,  by  a  plane  parallel  to  the  lateral  edges, 
is  a  parallelogram. 

51.  What  length  of  canvas  f  yd.  wide  is  required  to  make  a  conical 
tent  14  ft.  in  diameter  and  10  ft.  high? 

52.  In  a  cube  the  square  of  a  diagonal  is  three  times  the  square  of  an 
edge. 

53.  How  many  square  feet  of  tin  will  be  required  to  make  a  funnel,  if 
the  diameters  of  the  top  and  the  bottom  are  to  be  30  in.  and  15  in.  respec- 
tively, and  the  height  25  in.  ? 

54.  The  four  middle  points  of  two  pairs  of  opposite  edges  of  a  tetra- 
edron  are  in  one  plane,  and  at  the  vertices  of  a  parallelogram. 

55.  The  section  of  a  triangular  pyramid  made  by  a  plane  passed  parallel 
to  two  opposite  edges  is  a  parallelogram. 

56.  The  diameters  of  the  bases  of  a  frustum  of  a  cone  are  10  in.  and 
8  in.  respectively,  and  its  slant  height  is  14  in.     Find  its  lateral  area. 

57.  A  right  circular  cylinder  6  ft.  in  diameter  is  equivalent  to  a  right 
circular  cone  7  ft.  in  diameter.     If  the  height  of  the  cone  is  8  ft.,  what  is 
the  height  of  the  cylinder  ? 

58.  Find  the  surface  of  a  cubical  cistern  whose  contents  are  373,248 
cu.  in. 

59.  The  frustum  of  a  right  circular  cone  is  14  ft.  high,  and  has  a 
volume  of  924  cu.  ft.     Find  the  radii  of  its  bases  if  their  sum  is  9  ft. 

60.  The  plane  which  bisects  a  diedral  angle  of  a  tetraedron  divides  the 
opposite  side  into  segments,  which  are  proportional  to  the  areas  of  the 
adjacent  faces. 

61.   Find  the  area  of  a  section  in  Problem  59  equidistant  from  the 
bases. 


APPENDIX.  239 

62.  A  Dutch  windmill  in  the  shape  of  a  frustum  of  a  right  cone  is 
12  metres  high.     The  outer  diameters  at  the  bottom  and  the  top  are 
16  metres  and  12  metres,  the  inner  diameters  12  metres  and  10  metres, 
respectively.     How  many  cubic  metres  of  stone  were  required  to  build  it  ? 

63.  The  volume  of  a  truncated  parallelepiped  is  equal  to  the  product 
of  a  right  section  by  one-fourth  the  sum  of  its  four  lateral  edges. 

64.  The  Pyramid  of  Cheops  was  originally  480.75  ft.  high,  and  764  ft. 
square  at  the  base.     What  was  its  volume  ? 

65.  The  volume  of  a  cylinder  of  revolution  is  equal  to  the  product  of 
its  lateral  area  by  half  its  radius. 

66.  If  a  spherical  shell  have  an  exterior  diameter  of  14  in.,  what  should 
be  the  thickness  of  the  wall  so  that  it  may  contain  696.9  cu.  in.  ? 

67.  Find  the  depth  of  a  cubical  cistern  which  shall  hold  2000  gallons, 
each  gallon  being  231  cu.  in. 

68.  If  an  iron  sphere,  12  in.  in  diameter,  weighs  n  Ibs.,  what  will  be 
the  weight  of  an  iron  sphere  whose  diameter  is  16  in.  ? 

69.  Find  the  depth  of  a  cubical  box  which  shall  contain  100  bu.  oi: 
grain,  each  bushel  holding  2150.42  cu.  in. 

70.  A  cone  of  revolution  whose  vertical  angle  is  60°,  is  circumscribed 
about  a  sphere.     Compare  the  area  of  the  sphere  and  the  lateral  area  of 
the  cone.     Compare  their  volumes. 

71.  The  altitudes  of  two  similar  cones  of  revolution  are  as  11  to  9. 
What  is  the  ratio  of  their  total  areas  ?     Of  their  volumes  ? 


240  PLANE  GEOMETRY. 


ADDITIONAL   EXERCISES   ON   BOOK  VIII. 

1.  How  many  points  on  a  spherical  surface  determine  a  small  circle  ? 
How  many,  in  general,  determine  a  great  circle  ? 

2.  The  polar  triangle  of  a  trirectangular  triangle  is  a  trirectangular 
triangle  coinciding  with  the  triangle  itself. 

3.  Any  lune  is  to  a  trirectangular  triangle  as  its  angle  is  to  half  a 
right  angle. 

4.  If  the  radius  of  a  sphere  is  bisected  at  right  angles  by  a  plane,  the 
two  zones  into  which  the  surface  of  the  sphere  is  divided  are  to  each 
other  as  3:1. 

5.  If  the  radii  of  two  spheres  are  6  in.  and  4  in.  respectively,  and  the 
distance  between  their  centres  is  5  in.,  what  is  the  area  of  the  circle  of 
intersection  of  these  spheres  ? 

6.  Find  the  diameter  of  a  sphere  whose  volume  is  one  cubic  foot. 

7.  Find  the  area  of  a  spherical  triangle  each  of  whose  angles  is  70°, 
on  a  sphere  whose  surface  is  144  sq.  in. 

8.  Find  the  radius  of  the  circle  determined,  in  a  sphere  of  3  in. 
diameter,  by  a  plane  1  in.  from  the  centre. 

9.  Find  the  area  of  a  birectangular  triangle  whose  vertical  angle  is 
108°  on  a  sphere  whose  surface  is  400  sq.  in. 

10.  A  spherical  triangle  is  to  the  surface  of  the  sphere  as  the  spherical 
excess  is  to  eight  right  angles. 

11.  In  any  right  spherical  triangle,  if  one  side  be  greater  than  a  quad- 
rant, there  must  be  a  second  side  greater  than  a  quadrant. 

12.  Find  the  angles  of  an  equilateral  spherical  triangle  whose  area  is 
equal  to  that  of  a  great  circle. 

13.  Considering  the  moon  as  a  circle  of  diameter  2160.6  mi.,  whose 
centre  is  23,4820  mi.  from  the  eye,  what  is  the  volume  of  the  cone  whose 
base  is  the  full  moon  and  whose  vertex  is  the  eye  ? 

14.  One  sphere  has  twice  the  volume  of  another.     Find  the  ratio  of 
the  radius  of  the  first  to  the  radius  of  the  second. 


APPENDIX.  241 

15.  The  circumference  of  a  hemispherical  dome  is  132  ft.     How  many 
square  feet  of  lead  are  required  to  cover  it  ? 

16.  Find  the  surface  of  a  sphere  whose  volume  is  2  cu.  ft. 

17.  If  the  ball  on  the  top  of  St.  Paul's  Cathedral  in  London  is  6  ft.  in 
diameter,  what  would  it  cost  to  gild  it  at  7  cents  per  square  inch  ? 

18.  What  is  the  ratio  of  the  surface  of  a  sphere  to  the  entire  surface  of 
its  hemisphere  ? 

19.  Find  the  ratios  of  the  areas  of  two  spherical  triangles  on  the  same 
sphere,  the  angles  being  60°,  84°,  129°,  and  80°,  110°,  114°  respectively. 

20.  Prove  that  the  areas  of  zones  on  equal  spheres  are  proportional  to 
their  altitudes. 

21.  Find  a  circumference  of  a  small  circle  of  a  sphere  whose  diameter 
is  20  in.,  the  plane  of  the  circle  being  5  in.  from  the  centre  of  the  sphere. 

22.  The  diameter  of  a  sphere  is  21  ft.     Find  the  curved  surface  of 
a  segment  whose  height  is  6  ft. 

23.  Find  the  volume  of  a  sphere  inscribed  in  a  cube  whose  volume 
is  1331  cu.  in. 

24.  The  altitude  of  the  torrid  zone  is  about  3200  mi.    Find  its  area  in 
square  miles,  assuming  the  earth  to  be  a  sphere  with  a  radius  of  4000  mi. 

25.  A  spherical  pyramid  has  for  base  a  trirectangular  triangle.     What 
fraction  is  the  pyramid  of  the  sphere  ? 

26.  Find  the  ratio  of  a  spherical  surface  to  the  cylindrical  surface  of 
the  circumscribed  cylinder. 

27.  The  radii  of  two  concentric  spheres  are  8  and  12  in. ;  a  plane  is 
drawn  tangent  to  the  interior  sphere.     Find  the  area  of  the  section  made 
in  the  outer  sphere. 

28.  If  an  iron  ball  4  in.  in  diameter  weighs  9  Ibs.,  what  is  the  weight 
of  a  hollow  iron  shell  2  in.  thick,  whose  external  diameter  is  20  in.  ? 

29.  The  surface  of  a  sphere  is  to  be  800  sq.  in.     What  radius  should 
be  taken  ? 

30.  What  is  the  ratio  of  the  entire  surface  of  a  cylinder  circumscribed 
about  a  sphere  to  the  entire  surface  of  its  hemisphere  ? 

31.  To  construct  on  the   spherical   blackboard  a  spherical  triangle, 
having  a  side  75°,  and  the  adjacent  angles  110°  and  87°. 

32.  The  radius  of  the  base  of  the  segment  of  a  sphere  is  16  in.,  and  the 
radius  of  the  sphere  is  20  in.     Find  its  volume. 

33.  Two  spheres  have  radii  of  8  in.  and  7  in.  respectively.     What  is 
the  ratio  of  the  surfaces  of  those  spheres  ?     Of  their  volumes  ? 


242  SOLID    GEOMETRY. 

34.  A  cone  has  for  its  base  a  great  circle  of  a  sphere,  and  for  its  vertex 
a  pole  of  that  circle.     Find  the  ratio  of  the  curved  surfaces  of  the  cone 
and  hemisphere  ;  of  the  entire  surfaces. 

35.  To  draw  an  arc  of  a  great  circle  perpendicular  to  a  spherical  arc, 
from  a  given  point  without  it. 

36.  The  radius  of  the  base  of  a  segment  of  a  sphere  is  40  ft. ,  and  its 
height  is  20  ft. ;  find  its  volume. 

37.  Find  the  altitude  of  a  zone  whose  area  is  100  sq.  in.,  on  a  surface 
of  a  sphere  of  13  in.  radius. 

38.  The  area  of  a  zone  of  one  base  (the  other  base  is  zero)  equals  that 
of  a  circle  whose  radius  is  the  chord  of  the  generating  arc. 

39.  At  a  given  point  in  a  great  circle,  to  draw  an  arc  of  a  great  circle, 
making  a  given  angle  with  the  first. 

40.  The  surface  of  a  sphere  is  81  sq.  in.     Find  its  volume. 

41.  In  a  right-angled  spherical  triangle,  if  one  side  is  equal  to  a  quad- 
rant, so  is  another  side. 

42.  The  altitude  of  a  prism  is  9  ft.  and  the  perimeter  of  the  base  12  ft. ; 
find  the  altitude  and  perimeter  of  a  base  of  a  similar  prism  one-third  as 
great. 

43.  The  volume  of  a  sphere  is  7  cu.  ft.     Find  its  diameter  and  surface. 

44.  The  volume  of  a  spherical  sector  is  36  cu.  in.  ;  the  diameter  of  the 
sphere  is  18  in.     Find  the  area  of  the  zone  that  forms  the  base  of  the 
sector. 

45.  The  mean  radii  of  the  earth  and  moon  are  respectively  3956  mi.; 
1080.3  mi.     Show  that  their  volumes  are  as  49  to  1,  nearly. 

46.  If  lines  are  drawn  from  any  point  in  the  surface  of  a  sphere  to  the 
ends  of  a  diameter,  they  will  form  with  each  other  a  right  angle. 

47.  The  volumes  of  two  spheres  are  as  27  is  to  64.    Find  the  ratio 
(1)  of  their  diameters  ;  (2)  of  their  surfaces. 

48.  The  mean  diameter  of  the  planet  Jupiter  being  86,657  mi.,  find 
the  ratio  of  its  volume  to  that  of  the  earth. 

49.  If  two  straight  lines  are  tangent  to  a  sphere  at  the  same  point,  the 
plane  of  these  lines  is  tangent  to  the  sphere. 

50.  In  a  sphere  whose  radius  is  5  in.,  find  the  altitude  of  a  zone  whose 
area  shall  be  that  of  a  great  circle. 


APPENDIX.  243 

51.  The  sun's  diameter  is  about  109  times  the  earth's.     Find  the  ratio 
of  their  volumes. 

52.  Any  lime  is  to  a  trirectangular  triangle  as  its  angle  is  to  half  a 
right  angle. 

53.  What  is  the  radius  of  that  sphere  whose  number  of  square  units 
of  surface  equals  the  number  of  cubic  units  of  volume  ? 

54.  The  largest  possible  cube  is  cut  out  of  a  sphere  one  foot  in  diameter. 
Find  the  length  of  an  edge. 

55.  Given  a  sphere  of  radius  10.     How  far  from  the  centre  must  the 
eye  be  in  order  to  see  one-fourth  of  its  surface  ? 

56.  What  is  the  radius  of  that  sphere  whose  number  of  cubic  units  of 
volume  equals  the  number  of  square  units  of  area  in  one  of  its  great 
circles  ? 

57.  Spherical  polygons  are  to  each  other  as  their  spherical  excesses. 

58.  A  cone,  a  sphere,  and  a  cylinder  have  the  same  altitudes  and  diam- 
eters.    Show  that  their  volumes  are  in  arithmetical  progression. 

59.  If  the  angles  of  a  spherical  triangle  are  respectively  65°,  112°,  and 
85°,  how  many  degrees  are  there  in  each  side  of  its  polar  triangle  ? 

60.  A  metre  was  originally  intended  to  be  0.000  000  1  of  a  quadrant  of 
the  circumference  of  the  earth.     Assuming  it  to  be  such,  and  the  earth 
to  be  a  sphere,  find  (1)  its  radius  in  kilometers;  (2)  its  volume  in  cubic 
kilometers. 

61.  Given  the  spherical  triangle  whose  sides  are  respectively  80°,  90°, 
and  1-10°,  find  the  angles  of  its  polar  triangle. 

62.  If  the  atmosphere  extends  to  a  height  of  45  miles  above  the  earth's 
surface,  what  is  the  ratio  of  its  volume  to  the  volume  of  the  earth,  assum- 
ing the  latter  to  be  a  sphere  with  a  diameter  of  7912  mi.  ? 

63.  What  part  of  the  surface  of  a  sphere  is  a  lune  whose  angle  is  45°? 
54°  ?  80°  ? 


244  SOLID   GEOMETRY. 


MISCELLANEOUS   EXERCISES.     BOOKS   VI.-VHI. 

(SOLID.) 

1.  Find  the  lateral  area  of  a  right  pentagonal  pyramid  whose  slant 
height  is  9  in.,  and  each  side  of  the  base  6  in. 

2.  A  pyramid  20  ft.  high  has  a  base  containing  169  sq.  ft.     How  far 
from  the  vertex  must  a  plane  be  passed  parallel  to  the  base,  so  that  the 
section  may  contain  100  sq.  ft.  ? 

3.  A  pyramid  16  ft.  high  has  a  square  base  10  ft.  on  a  side.     Find  the 
area  of  a  section  made  by  a  plane  parallel  to  the  base  and  6  ft.  from  the 
vertex. 

4.  The  volume  of  the  frustum  of   a  regular  hexagonal  pyramid  is 
12  cu.  ft.,  the  sides  of  the  bases  are  2  ft.  and  1  ft.;  find  the  height  of  the 
frustum. 

5.  A  regular  pyramid  8  ft.  high  is  transformed  into  a  regular  prism 
with  an  equivalent  base  ;  find  the  height  of  the  prism. 

6.  The  lateral  area  of  a  cylinder  of  revolution  is  equal  to  the  area  of 
a  circle  whose  radius  is  a  mean  proportional  between  the  altitude  of  the 
cylinder  and  the  diameter  of  its  base. 

7.  The  lateral  area  of  a  given  cone  of  revolution  is  double  the  area  of 
its  base  ;  find  the  ratio  of  its  altitude  to  the  radius  of  its  base. 

8.  Find  the  volume  of  the  frustum  of  a  cone  of  revolution,  the  radii 
of  the  bases  being  8  ft.  and  4  ft.  and  the  altitude  6  ft. 

9.  The  projections  of  parallel  straight  lines  on  any  plane  are  them- 
selves parallel. 

10.  Construct  an  equilateral  triangle  equal  to  a  given  triangle. 

11.  Construct  an  isosceles  triangle  having  each  angle  at  the  base  double 
the  third  angle. 

12.  The  total  area  of  a  cone  of  revolution  is  500  sq.  in.  ;  its  altitude  is 
10  in.     What  is  the  diameter  of  its  base  ? 

13.  What  is  the  lateral  area  and  the  total  area  of  a  frustum  of  a  cone 
of  revolution  whose  altitude  is  30  in.,  and  the  diameters  of  whose  bases 
are  9  in.  and  21  in.  respectively  ? 


APPENDIX.  245 

14.  The  diameters  of  the  bases  of  a  frustum  of  a  cone  of  revolution  are 
1\  in.  and  12  in.  respectively  ;  its  volume  is  575  cu.  in.    What  is  its  altitude  ? 

15.  If  the  altitude  of  a  cylinder  of  revolution  is  equal  to  the  diameter 
of  its  base,  the  volume  is  equal  to  the  product  of  its  total  area  by  one- 
third  of  its  radius. 

16.  Find  the  ratio  of  two  rectangular  parallelepipeds,  if  their  dimen- 
sions are  4,  7,  9,  and  8,  14,  18  respectively. 

17.  The  volume  of  a  sphere  is  one  cubic  foot.     Find  the  surface  of  the 
circumscribing  cylinder. 

18.  How  far  from  the  base  must  a  cone,  whose  altitude  is  64  in.,  be 
cut  by  a  plane  so  that  the  frustum  shall  be  equivalent  to  half  the  cone? 

19.  What  should  be  the  altitude  of  a  cone  of  revolution  whose  base  has 
a  diameter  of  15  in.,  so  that  the  lateral  area  may  be  20  square  feet  ? 

20.  The  altitude  of  a  cone  of  revolution  is  four  times  the  radius  of  its 
base  ;  the  lateral  area  is  1000  sq.  in.     Find  the  radius  and  altitude. 

21.  The  total  area  of  a  cylinder  of  revolution  is  800  sq.  in. ;  its  altitude 
is  16  in.     What  is  the  diameter  of  a  base  ? 

22.  What  should  be  the  volume  of   a  cylinder  of   revolution  whose 
altitude  is  20  in.,  so  that  the  lateral  area  shall  be  3  sq.  ft.  ? 

23.  Two  given  straight  lines  do  not  intersect  and  are  not  parallel. 
Find  a  plane  on  which  their  projections  will  be  parallel. 

24.  Describe  a  circle  which  shall  touch  a  given  circle  and  two  given 
straight  lines  which  themselves  touch  the  given  circle. 

25.  Pass  a  plane  perpendicular  to  a  given  straight  line  through  a  given 
point  not  in  that  line. 

26.  In  any  triedral  angle,  the  planes  bisecting  the  three  diedral  angles 
all  intersect  in  the  same  straight  line. 

27.  Draw  a  straight  line  through  a  given  point  in  space,  so  that  it  shall 
cut  two  given  straight  lines  not  in  the  same  plane. 

28.  Find  the  dimensions  of  a  cube  whose  surface  is  numerically  equal 
to  its  contents. 

29.  The  base  of  a  regular  pyramid  is  a  hexagon  whose  side  is  12  ft. 
Find  the  height  of  the  pyramid  if  the  lateral  area  is  eight  times  the  area 
of  the  base. 

30.  Find  the  volume  of  the  frustum  of  a  regular  triangular  pyramid, 
the  sides  of  whose  bases  are  18  and  16,  and  whose  lateral  edge  is  11. 


246  SOLID    GEOMETRY. 

31.  Find  the  lateral  area  of  a  right  pyramid  whose  slant  height  is  8  ft., 
and  whose  base  is  a  regular  octagon  of  which  each  side  is  6  ft.  long. 

32.  Find  the  volume  of  the  frustum  of  a  square  pyramid,  the  sides  of 
whose  bases  are  16  and  12  ft.,  and  whose  altitude  is  24  ft. 

33.  The  altitudes  of  two  similar  cylinders  of  revolution  are  as  6  to  5. 
What  is  the  ratio  of  their  total  areas  ?     Of  their  volumes  ? 

34.  Find  the  ratio  of  two  rectangular  parallelepipeds,  if  their  altitudes 
are  each  6  ft.,  and  their  bases  8  ft.  by  4  ft.,  and  15  ft.  by  10  ft.  respectively. 

35.  In  order  that  a  cylindrical  tank  with  a  depth  of  24  ft.  may  contain 
2000  gal.,  what  should  be  its  diameter  ? 

36.  How  many  cubic  inches  of  iron  would  be  required  to  make  that 
tank,  its  walls  being  one-fourth  of  an  inch  thick  ? 

37.  The  diameter  of  a  right  circular  cylinder  is  12  ft.,  and  its  altitude 
9  ft.     What  is  the  side  of  an  equivalent  cube  ? 

38.  A  sphere  6  in.  in  diameter  has  a  hole  bored  through  its  centre  with 
a  2-inch  auger  ;  find  the  remaining  volume. 

39.  How  high  above  the  earth  must  a  person  be  raised  in  order  that  he 
may  see  one-fifth  of  its  surface  ? 

40.  How  much  of  the  earth's  surface  would  a  man  see  if  he  were  raised 
to  the  height  of  the  diameter  above  it  ? 

41.  Find  the  volume  of  a  spherical  segment  of  one  base  whose  altitude 
is  4  ft. ,  the  radius  of  the  sphere  being  10  ft. 

42.  Find  the  surface  of  a  sphere  inscribed  in  a  cube  whose  surface 
is  216. 

43.  The  volumes  of  two  similar  cones  of  revolution  are  to  each  other 
as  3  :  5  ;  find  the  ratio  of  their  lateral  areas,  and  of  their  volumes. 

44.  The  lateral  area  of  a  cone  of  revolution  is  60  TT  and  its  slant  height 
is  16 ;  find  its  volume. 

45.  Find  the  lateral  area  of  the  frustum  of  a  cone  of  revolution,  the 
radii  of  the  bases  being  42  and  12  in.,  and  the  altitude  36  in. 

46.  The  two  legs  of  a  right  triangle  are  a  and  b  ;  find  the  area  of  the 
surface  generated  when  the  triangle  revolves  about  its  hypotenuse. 

47.  Find  the  volume  of   a  regular  icosaedron  whose  edges  are  each 
20  ft. 


APPENDIX.  247 

48.  The  base  of  a  cone  is  equal  to  a  great  circle  of  a  sphere,  and  the 
altitude  of  the  cone  is  equal  to  a  diameter  of  the  sphere ;  compare  the 
volumes  of  the  cone  and  the  sphere. 

49.  The  lateral  area  of  a  cylinder  of  revolution  is  116|  sq.  ft.,  and  the 
altitude  is  14  ft.  ;  find  the  diameter  of  its  base. 

50.  Find  the  number  of  cubic  feet  in  the  trunk  of  a  tree,  70  ft.  long, 
the  diameters  of  its  ends  being  9  and  7  ft. 

51.  The  heights  of  two  cylinders  of  revolution  of  equal  volumes  are  as 
9  : 16  ;  the  diameter  of  one  is  6  ft.     Find  the  diameter  of  the  other. 

52.  The  volume  of  a  sphere  is  113  ;  find  its  diameter  and  its  surface. 

53.  The  volume  of  a  sphere  is  776  TT  ;  find  its  diameter  and  its  surface. 

54.  Find  the  weight  of  an  iron  shell  4  in.  in  diameter,  the  iron  being 
1  in.  thick,  and  weighing  |  Ib.  to  the  cubic  inch. 

55.  If  an  iron  ball  8  in.  in  diameter  weighs  72  Ibs.,  find  the  weight  of 
an  iron  shell  10  in.  in  diameter,  the  iron  being  2  in.  thick. 

56.  A  sphere,  2  ft.  in  diameter,  is  cut  by  two  parallel  planes,  one  at  3 
and  the  other  at  9  in.  from  the  centre  ;  find  the  volume  of  the  segment 
included  between  them. 

57.  If  the  angle  of  a  lune  is  50°,  find  its  area  on  a  sphere  whose  surface 

is  72  sq.  in. 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

LOAN  DEPT. 

This  book  is  due  on  the  last  date  stamped  below,  or 
on  the  date  to  which  renewed. 

Renewed  books  are  subject  to  immediate  recall. 


*-~D   rrn 

fjftfU    *   0          -I.HJ 

TOflrrS  381 

I3*rf®w 

SEC  D  LD 

MQV  i  5  1962 

LD  21A-50m-4,'60 
(A9562slO)476B 


General  Library 

University  of  California 

Berkeley 


9183G6 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


